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Home , Contour integration, Lebesgue integration, Measure (mathematics), Outer measure, Vitali set

Measure Theory and

T. Muthukumar [email protected]

April 12, 2018 ii Contents

Notationsv

1 Introduction1 1.1 Riemann Integration and its Inadequacy...... 1 1.1.1 and : Interchange...... 3 1.1.2 Differentiation and Integration: Duality...... 5 1.2 Motivating Lebesgue Integral and ...... 6

2 on Rn 9 2.1 Introduction...... 9 2.2 ...... 12 2.2.1 Abstract -up...... 22 2.3 Measurable Sets...... 22 2.3.1 Abstract Set-up...... 32 2.4 Measurable Functions...... 34 2.5 Littlewood’s Three Principles...... 43 2.5.1 First Principle...... 44 2.5.2 Third Principle...... 45 2.5.3 Second Principle...... 49 2.6 Jordan Content or Measure...... 50

3 Lebesgue Integration 55 3.1 Simple Functions...... 55 3.2 Bounded With ...... 58 3.3 Non-negative Functions...... 62 3.4 General Integrable Functions...... 67 3.5 Order of Integration...... 75 3.6 Lp Spaces...... 78

iii CONTENTS iv

3.7 Invariance of Lebesgue Integral...... 87

4 Duality of Differentiation and Integration 89 4.1 Monotone Functions...... 90 4.2 Functions...... 97 4.3 of an Integral...... 106 4.4 and FTC...... 111

Appendices 119

A and 121

Bibliography 127 Notations

Symbols

2S will denote the , the set of all , of a set S

L(Rn) the class of all Lebesgue measurable subsets of Rn

C denotes the plane of complex numbers

Q denotes the set of all rationals in R

Qn set of all vectors in Rn with each coordinate being

R denotes the

Rn denotes the Euclidean of dimension n Function Spaces

R([a, b]) denotes the space of all Riemann integrable functions on the [a, b]

Lip(E) denotes the space of all Lipschitz functions on E

AC(E) denotes the space of all absolutely continuous functions on E

BV (E) denotes the space of all bounded variation functions on E

C(X) the class of all real-valued continuous functions on X

C0(X) denotes the space of all continuous functions vanishing at ∞ on X

Cc(X) denotes the space of all compactly supported continuous functions on X

v NOTATIONS vi

Lp(E) denotes the space of all measurable p-integrable functions on E

M(Rn) the class of all finite a.e. real valued Lebesgue measurable functions on Rn General Conventions

Br(x) will denote the closed ball of radius r and centre at x Ec will denote the set of E ⊂ S, S \ E Chapter 1

Introduction

1.1 Riemann Integration and its Inadequacy

Let f :[a, b] → R be a bounded function. Let P be the partition of the interval [a, b], a = x0 ≤ x1 ≤ ... ≤ xk = b. For i = 0, 1, 2, . . . , k, let

Mi(P ) = sup f(x) and mi(P ) = inf f(x). x∈[x ,x ] x∈[xi−1,xi] i−1 i The upper of f with respect to the given partition P is,

k X U(P, f) = Mi(P )(xi − xi−1) i=1 and the lower Riemann sum of f with respect to the given partition P is,

k X L(P, f) = mi(P )(xi − xi−1). i=1 We say the bounded function f is Riemann integrable on [a, b] if the infimum of upper sum and supremum of lower sum, over all partitions P of [a, b], coincide and is denoted as Z b f(x) dx := inf U(P, f) = sup L(P, f). a P P If f = u + iv is a bounded complex-valued function on [a, b], then f is said to be Riemann integrable if its real and imaginary parts are Riemann

1 CHAPTER 1. INTRODUCTION 2 integral and Z b Z b Z b f(x) dx = u(x) dx + i v(x) dx. a a a If either f is unbounded or the domain [a, b] is not finite then its corre- sponding integral, called as , is defined in terms of limits of Riemann integrable functions, whenever possible. Exercise 1. Every Riemann integrable function1 is bounded. Let R([a, b]) denote the space of all Riemann integrable functions on [a, b]. The space R([a, b]) forms a over R (or C). It is closed under composition, if it makes sense.

Theorem 1.1.1. If f is continuous on [a, b], then f ∈ R([a, b]).

In fact even piecewise continuity is sufficient for Riemann integrability.

Theorem 1.1.2. If f is continuous except at finitely many points of [a, b] (piecewise continu- ous), then f ∈ R([a, b]).

But there are functions which has discontinuity at countably many points and are still in R([a, b]). Example 1.1. Consider the function

 1 1 1 if k+1 < x ≤ k and k is odd  1 1 f(x) = 0 if k+1 < x ≤ k and k is even 0 x = 0 which has discontinuities at x = 0 and x = 1/k, for k = 1, 2,.... It can be shown that f ∈ R([0, 1]).

Theorem 1.1.3. If f is bounded monotonic on [a, b] then f ∈ R([a, b]).

In fact, one can construct functions whose set of discontinuities are ‘dense’ in [0, 1].

1here by Riemann integrable we mean the upper sum and lower sum coincide and are finite CHAPTER 1. INTRODUCTION 3

∞ Example 1.2. For instance, let {rk}1 denote a countable dense of [0, 1] (for instance, Q) and define

∞ X 1 f(x) = H(x − r ) k2 k k=1 where H : R → R is defined as ( 1 if x ≥ 0 H(x) = 0 if x < 0.

The function f is discontinuous at all the points rk and can be shown to be in R([0, 1]), because it is bounded and monotone.

Theorem 1.1.4. If f ∈ R([a, b]) then f is continuous on a dense subset of [a, b].

Example 1.3. An example of a function f : [0, 1] → R which is not Riemann integrable is ( 1 x ∈ f(x) = Q 0 x ∈ [0, 1] \ Q. A necessary and sufficient condition of Riemann integrability is given by Theorem 3.0.1. Thus, even to characterise the class of Riemann integrable functions, we need to have the notion of (“measure”) (at least measure zero).

1.1.1 Limit and Integral: Interchange

Let us consider a of functions {fk} ⊂ R([a, b]) and define f(x) := limk→∞ fk(x), assuming that the limit exists for every x ∈ [a, b]. Does f ∈ R([a, b])? The answer is a “no”, as seen in example below. Example 1.4. Fix an enumeration (order) of the set of rationals in [0, 1]. Let the finite set rk denote the first k elements of the set of rationals in [0, 1]. Define the sequence of functions ( 1 if x ∈ rk fk(x) = 0 otherwise. CHAPTER 1. INTRODUCTION 4

Each fk ∈ R([0, 1]), since it has discontinuity at k (finite) number of points. The point-wise limit of fk, f = limk→∞ fk, is ( 1 x ∈ f(x) = Q 0 x ∈ [0, 1] \ Q which we have seen above is not Riemann integrable. Thus, the space R([a, b]) is not “complete” under point-wise limit. How- ever, R([a, b]) is complete under . A related question is if the limit f ∈ R([a, b]), is the of f the limit of the Riemann of fk, i.e., can we say

Z b Z b f(x) dx = lim fk(x) dx? a k→∞ a The answer is a “no” again. Example 1.5. Consider the functions ( k x ∈ (0, 1/k) fk(x) = 0 otherwise

Then f(x) = limk fk(x) = 0. Note that Z fk(x) dx = 1 ∀k, R but R f(x) dx = 0. R The interchange becomes possible under uniform convergence.

Theorem 1.1.5. Let {fk} ⊂ R([a, b]) and fk(x) → f(x) uniformly in [a, b]. Then f ∈ R([a, b]) and Z b Z b f = lim fk. a k→∞ a But uniform convergence is too demanding in practice. The following more general result for interchanging limit and integral will be proved in this write-up. CHAPTER 1. INTRODUCTION 5

Theorem 1.1.6. Let {fk} ⊂ R([a, b]) and f ∈ R([a, b]). Also, let fk(x) → f(x) point-wise and fk are uniformly bounded. Then Z b Z b lim fk = f. k→∞ a a The proof of above theorem is not elementary, thus in classical analysis we always prove the result for uniform convergence. Observe the hypothesis of integrability on f in the above theorem.

1.1.2 Differentiation and Integration: Duality An observation we make, once we have Riemann integration, is about the dual nature of differentiation and integration. Thus, one asks the following two questions: 1. (Derivative of an integral) For which class of functions can we say d Z x f(t) dt = f(x)? dx a

2. (Integral of a derivative) For which class of functions can we say Z b f 0(x) dx = f(b) − f(a)? a

To answer the first question, for any f ∈ R([a, b]), let us define the function Z x F (x) := f(t) dt. a Exercise 2. Show that if f ∈ R([a, b]) then F is continuous on [a, b]. The first question is answered by the following result of Riemann inte- gration. Theorem 1.1.7. Let f ∈ R([a, b]). If f is continuous at a point x ∈ [a, b], then F is differentiable at x and F 0(x) = f(x). What is the most general class of functions for which the above result holds true. The second question is answered by the famous Fundamental theorem of (FTC). CHAPTER 1. INTRODUCTION 6

Theorem 1.1.8 (Fundamental Theorem of Calculus). If f is differentiable function on [a, b] such that f 0 ∈ R([a, b]), then

Z b f 0(x) dx = f(b) − f(a). a Note that the fundamental theorem of calculus fails under the following two circumstances:

1. For a f on [a, b] which is nowhere differentiable on [a, b]. Do such functions exist?

2. Derivative of f exists for all points in [a, b], but f 0 is not integrable. Do such functions exist?

K. Weierstrass was the first to show in 1872 the existence of a everywhere continuous function which is nowhere differentiable. Prior to Weierstrass’ proof it was believed that every continuous function is differentiable except on a set of “isolated” points. This example of Weierstrass showed the existence of function for which FTC may not make any sense. The existence of a function f whose derivative exists everywhere but the derivative is not integrable, was shown by Vito Volterra, who was a student of Ulisse Dini, in 1881. His example was a clever modification of the function ( x2 sin( 1 ) x 6= 0 g(x) = x 0 x = 0 which is differentiable. The derivative of g, g0(x) = 2x sin(1/x) − cos(1/x), is discontinuous at x = 0. A natural question to ask was: Identify the class of functions for which FTC makes sense.

1.2 Motivating Lebesgue Integral and Mea- sure

Whatever the reasons are, we should be convinced now that it is worthwhile looking for a new type of integration which coincides for Riemann integrable functions and also includes “non-integrable” (Riemann) functions. CHAPTER 1. INTRODUCTION 7

The Riemann integration was based on the simple fact that one can in- tegrate step functions (piecewise constant) and then approximate any given function with piecewise constant functions, by partitioning the domain of the function. Lebesgue came up with this idea of partitioning the range of the function. A very good analogy to motivate Lebesgue integration is the following (cf. [Pug04]): Suppose A asks both B and C to give the total value for a bunch of coins with all denominations lying on a table. First B counts them as he picks the coins and adds their denomination to come up with the total value. This is Riemann’s way of integration (partitioning the domain, if you consider the function to be coin mapped to its denomination). In his/her turn, C sorts the coin as per their denominations in to separate piles and counts the coins in each pile, multiply it with the denomination of the pile and sum them up for the total value. Both B and C will come up with the same value (assuming they counted right!). The way C counted is Lebesgue’s way of integration. We know that integration is related with the question of computing length// of a subset of an , depending on its di- mension. Now, if one wants to partition the range of a function, we need some way of “measuring” how much of the domain is sent to a particular region of the partition. This problem leads us to the theory of measures where we try to give a notion of “measure” to subsets of an Euclidean space. CHAPTER 1. INTRODUCTION 8 Chapter 2 n Lebesgue Measure on R

2.1 Introduction

In this chapter we shall develop the notion of Lebegue ‘measure’ in Rn.

Definition 2.1.1. We say R is a cell (open) in Rn if R is of the form (a1, b1) × (a2, b2) × ... × (an, bn), i.e.,

n n R = Πi=1(ai, bi) := {x = (x1, . . . , xn) ∈ R | xi ∈ (ai, bi) for all 1 ≤ i ≤ n}.

The volume (finite) of the cell R, denoted as |R|, is the non-negative number, n n |R| = Πi=1(bi − ai). We say R is closed if R = Πi=1[ai, bi].

We do not let ai = −∞ or bi = +∞, i.e., by a cell we always refer to a cell with finite volume. In fact, by our definition, cells in Rn are precisely those rectangles with finite volume in Rn whose sides are parallel to the coordinate axes. Also, a cell could refer to Cartesian products of open, closed or half- open or half-closed intervals. Note that if ai = bi, for all i, then we have the volume of an to be zero. Moreover, if R is the closure of an open cell R, then |R| = |R|. If ai = bi, for some i in a closed cell, then the (lower dimensional) cell also has volume zero. Exercise 3. Show that the volume of a cell is translation invariant, i.e., for any cell R, |R + x| = |R| for all x ∈ Rn. The volume is also dilation invariant, Qn i.e., for any λ = (λ1, . . . , λn) such that λi > 0, |λR| = i=1 λi|R|, where λR = {(λixi) | (xi) ∈ R}.

9 N CHAPTER 2. LEBESGUE MEASURE ON R 10

Exercise 4 (uniqueness of volume). Let C be the collection of all cells of Rn and if ν : C → [0, +∞) is a well defined set-function on C such that ν is invariant under translation and dilation. Show that ν is same as the volume |·| up to a constant, i.e., there exists a constant α ≥ 0 such that ν(R) = α|R| for all cells R ∈ C. In particular, if we additionally impose the condition that ν([0, 1]n) = 1 then ν(R) = |R| for all R ∈ C. Exercise 5. Show that the volume satisfies monotonicity, i.e., if R ⊂ Q, then |R| ≤ |Q|. k n k Exercise 6. If {Ri}1 are cells in R such that R ⊂ ∪i=1Ri, then |R| ≤ Pk i=1 |Ri|. Theorem 2.1.2. For every open subset Ω ⊂ R, there exists a unique count- ∞ able family of open intervals Ii such that Ω = ∪i=1Ii where Ii’s are pairwise disjoint. Proof. Since Ω is open, for every x ∈ Ω, there is an open interval in Ω that contains x. Let us pick the largest such open interval in Ω that contains x. How do we do this? Let, for each x ∈ Ω,

ax := inf {(a, x) ⊂ Ω} and bx := sup{(x, b) ⊂ Ω}. ax

Of course, ax and bx can take ±∞. Note that ax < x < bx. Set Ix := (ax, bx), is the largest open interval in Ω containing x. Thus, we have Ω = ∪x∈ΩIx. We shall now note that for any x, y ∈ Ω such that x 6= y, either Ix = Iy or Ix ∩ Iy = ∅. Suppose Ix ∩ Iy 6= ∅ then Ix ∪ Iy is also an open interval in Ω that contains x. Therefore, by the maximality of Ix, Ix ∪ Iy ⊂ Ix. Hence, Ix = Ix ∪ Iy. Similarly, Ix ∪ Iy = Iy. Thus, Ix = Iy and Ω is a disjoint of open intervals. It now only remains to show that the union can be made countable. Note that every open interval Ix contains a rational number. Since different intervals are disjoint, we can pick distinct rationals from each interval. Since rationals are countable, the collection of disjoint intervals cannot be uncountable. Thus, we have a countable collection of ∞ disjoint open intervals Ii such that Ω = ∪i=1Ii. From the uniqueness in the result proved above we are motivated to define the “length” of an open set Ω ⊂ R as the sum of the of the intervals n Ii. But this result has no exact analogue in R , for n ≥ 2. Exercise 7. An open connected set Ω ⊂ Rn, n ≥ 2 is the disjoint union of open cells iff Ω is itself an open cell. N CHAPTER 2. LEBESGUE MEASURE ON R 11

Exercise 8. Show that an open disc in R2 cannot be the disjoint union of open cells. However, relaxing our requirement to almost disjoint-ness (defined below) will generalise Theorem 2.1.2 to higher dimensions Rn, n ≥ 2.

Definition 2.1.3. We say a collection of cells Ri to be almost disjoint if the interiors of Ri are pairwise disjoint.

k Exercise 9. If a cell R = ∪i=1Ri such that Ri are pairwise almost disjoint, Pk then |R| = i=1 |Ri|. Theorem 2.1.4. For every open subset Ω ⊂ Rn, there exists a countable ∞ family of almost disjoint closed cells Ri such that Ω = ∪i=1Ri. Proof. To begin we consider the grid of cells in Rn of side length 1 and whose vertices have coordinates. The number cells in the grid is countable and they are almost disjoint. We ignore all those cells which are contained in Ωc. Now we have two families of cells, those which are contained in Ω, call the collection C, and those which intersect both Ω and Ωc. We bisect the latter cells further in to 2n cells of side each 1/2. Again ignore those contained in Ωc and add those contained in Ω to the collection C. Further bisecting the common cells in to cells of side length 1/4. Repeating this procedure, we have a countable collection C of almost disjoint cells in Ω. By k construction, ∪R∈CR ⊂ Ω. Let x ∈ Ω then there is a cell of side length 1/2 (after bisecting k times) in C which contains x. Thus, ∪R∈CR = Ω. Again, as we did in one dimension, we hope to define the “volume” of an open subset Ω ⊂ Rn as the sum of the of the cells R obtained in above theorem. However, since the collection of cells is not unique, in contrast to one dimension, it is not clear if the sum of the volumes is independent of the choice of your family of cells. We wish to extend the notion of volume to arbitrary subsets of an Eu- clidean space such that they coincide with the usual notion of volume for a cell, most importantly, preserving the properties of the volume. So, what are these properties of volume we wish to preserve? To state them, let’s first regard the volume as a on the power set of Rn, mapping to a non-negative . Thus, we wish to construct a ‘measure’ µ, n µ : 2R → [0, ∞] such that

1. If R is any cell of Rn, then µ(R) = |R|. N CHAPTER 2. LEBESGUE MEASURE ON R 12

2.( Translation Invariance) For every E ⊂ Rn, µ(E + x) = µ(E) for all x ∈ Rn. 3.( Monotonicity) If E ⊂ F , then µ(E) ≤ µ(F ).

∞ P∞ 4.( Countable Sub-additivity) If E = ∪i=1Ei then µ(E) ≤ i=1 µ(Ei).

∞ 5.( Countable Additivity) If E = ∪i=1Ei such that Ei are pairwise P∞ disjoint then µ(E) = i=1 µ(Ei). Exercise 10. Show that if µ obeys finite additivity and is non-negative, then µ is monotone. (Basically monotonicity is redundant from countable addi- tivity).

2.2 Outer measure

To construct a ‘measure’ on the power set of Rn, we use the simple approach of ‘covering’ an arbitrary subset of Rn by cells and assigning a unique number using them.

Definition 2.2.1. Let E ⊆ Rn, a subset of Rn. We say that a family of cells {Ri}i∈I is a cover of E iff E ⊆ ∪i∈I Ri. If each of the cell Ri in the cover is an open (resp. closed) cell, then the cover is said to be open (resp. closed) cover of E. If the index set I is finite/countable/uncountable, then the cover is said to be a finite/countable/uncountable cover.

We shall not consider the case of uncountable cover in this text, because uncountable additivity makes no sense. Exercise 11. Every subset of Rn admits a countable covering! n If we wish to associate a unique positive number to E ∈ 2R , satisfying monotonicity and (finite/countable) sub-additivity, then the association must satisfy

µ(E) ≤ µ (∪i∈I Ri) (due to monotonicity) X ≤ µ(Ri) (due to finite/countable sub-additivity) i∈I X = |Ri| (measure same as volume). i∈I N CHAPTER 2. LEBESGUE MEASURE ON R 13

The case when the index set I is strictly finite corresponds to Riemann integration which we wish to generalise. Thus, we let I to be a countable index set, henceforth. A brief note on the case when index set I is strictly finite is given in § 2.6.

Definition 2.2.2. For a subset E of Rn, we define its Lebesgue outer mea- sure1 µ?(E) as, ? X µ (E) := inf |Ri|, E⊆∪i∈I Ri i∈I the infimum being taken over all possible countable coverings of E.

The Lebesgue outer measure is a well-defined non-negative set function n on the power set of Rn, 2R . Exercise 12. The outer measure is unchanging if we restrict ourselves to open covering or closed covering, i.e., for every subset E ⊂ Rn,

? X µ (E) = inf |Si|, E⊆∪i∈I Si i∈I where the infimum is taken over all possible countable closed or open cover- ings {Si} of E. Before we see some examples for calculating outer measures of subsets of Rn, let us observe some immediate properties of outer measure following from definition.

Lemma 2.2.3. The outer measure µ? has the following properties:

(a) For every subset E ⊆ Rn, 0 ≤ µ?(E) ≤ +∞.

(b)( Translation Invariance) For every E ⊂ Rn, µ?(E + x) = µ?(E) for all x ∈ Rn. (c)( Monotone) If E ⊂ F , then µ?(E) ≤ µ?(F ).

∞ (d)( Countable Sub-additivity) If E = ∪i=1Ei then

∞ ? X ? µ (E) ≤ µ (Ei). i=1 1Why we call it “outer” and superscript with a ? will be clear in the next section N CHAPTER 2. LEBESGUE MEASURE ON R 14

Proof. (a) The non-negativity of the outer measure is an obvious conse- quence of the non-negativity of | · |.

(b) The invariance under translation is obvious too, by noting that for each covering {Ri} of E or E + x, {Ri + x} and {Ri − x} is a covering of E + x and E, respectively, and the volumes of the cell is invariant under translation.

(c) Monotonicity is obvious, by noting the fact that, the family of covering of F is a sub-family of the coverings of E. Thus, the infimum over the family of cover for E is smaller than the sub-family.

? (d) If µ (Ei) = +∞, for some i, then the result is trivially true. Thus, we ? assume that µ (Ei) < +∞, for all i. By the definition of outer measure, i ∞ for each ε > 0, there is a covering by cells {Rj}j=1 for Ei such that

∞ X ε |Ri | ≤ µ?(E ) + . j i 2i j=1

∞ i Since E = ∪i=1Ei, the family {Rj}i,j is a covering for E. Thus,

∞ ∞ ∞ ∞ X X X  ε  X µ?(E) ≤ |Ri | ≤ µ?(E ) + = µ?(E ) + ε. j i 2i i i=1 j=1 i=1 i=1

Since choice of ε is arbitrary, we have the countable sub-additivity of µ?.

We have seen the properties of outer measure. Let us now compute the outer measure for some subsets of Rn. Example 2.1. Outer measure of the empty set is zero, µ?(∅) = 0. Every cell is a cover for the empty set. Thus, infimum over the volume of all cells is zero. Example 2.2. The outer measure for a singleton set {x} in Rn is zero. The same argument as for empty set holds except that now the infimum is taken over all cells containing x. Thus, for each ε > 0, one can find a cell Rε such ? that x ∈ Rε and |Rε| ≤ ε. Therefore, µ ({x}) ≤ ε for all ε > 0 and hence µ?({x}) = 0. N CHAPTER 2. LEBESGUE MEASURE ON R 15

Example 2.3. The outer measure of any countable subset E of Rn is zero. A E = ∪x∈E{x}, where the union is countable. Thus, by countable sub-additivity, µ?(E) ≤ 0 and hence µ?(E) = 0. Let’s highlight something interesting at this stage. Note that the set of all rationals, Q, in R is countable. Hence µ?(Q) = 0. Also Q is dense in R. Thus, we actually have a dense (‘scattered’) subset of R whose outer measure is zero (‘small’). Example 2.4. The situation is even worse. The converse of above example is not true, i.e., we can have a whose outer measure is zero. The outer measure of Rn−1 (lower dimensional), for n ≥ 2, as a subset of Rn ? n−1 n−1 n−1 is zero, i.e., µ (R ) = 0. Choose a cover {Ri} of R in R such that −ε ε  n−1 n |Ri|n−1 = 1. Then Ei = Ri × 2i , 2i forms a cover for R in R . Thus,

∞ ? n−1 X µ (R ) ≤ |Ei| i=1 ∞ X 1 = 2ε = 2ε. 2i i=1

Since the choice of ε > 0 could be as small as possible, we have µ?(Rn−1) = 0. Example 2.5. Is there an uncountable subset of R whose outer measure zero? Consider the Cantor set C (cf. AppendixA) which is uncountable. Let us compute the outer measure of C. Recall that, for each i, Ci is disjoint union i −i ? i of 2 closed intervals, each of whose length is 3 . Thus, µ (Ci) = (2/3) , ? for all i. By construction, C ⊂ Ci and hence due to monotonicity µ (C) ≤ ? i i i→∞ ? µ (Ci) = (2/3) , for all i. But (2/3) −→ 0. Thus, µ (C) = 0. Example 2.6. A similar argument as above shows that the outer measure of the generalised Cantor set C (cf. AppendixA) is bounded above by

? k µ (C) ≤ lim 2 a1a2 . . . ak. k We shall, in fact, show that equality holds here using “continuity from above” of outer measure (cf. Example 2.12). Example 2.7. If R is any cell of Rn, then µ?(R) = |R|. Since R is a cover by itself, we have from definition, µ?(R) ≤ |R|. It now remains to prove the ∞ reverse . Let S be a closed cell. Let {Ri}1 be an arbitrary covering of S. Choose an arbitrary ε > 0. For each i, we choose an open cell Qi such that Ri ⊂ Qi and |Qi| < |Ri| + ε|Ri|. Note that {Qi} is an open covering of N CHAPTER 2. LEBESGUE MEASURE ON R 16

S. Since S is compact in Rn (closed and bounded), we can extract a finite k sub-cover such that S ⊂ ∪i=1Qi. Therefore,

k X |S| ≤ |Qi| (cf. Exercise6) i=1 k X ≤ (1 + ε) |Ri|. i=1 Since ε can be chosen as small as possible, we have

k ∞ X X |S| ≤ |Ri| ≤ |Ri|. i=1 i=1

Since {Ri} was an arbitrary choice of cover for S, taking infimum, we get |S| ≤ µ?(S). Thus, for a closed cell we have shown that |S| = µ?(S). Now, for the given cell R and ε > 0, one can always choose a closed cell S ⊂ R such that |R| < |S| + ε. Thus,

|R| < |S| + ε = µ?(S) + ε ≤ µ?(R) + ε (by monotonicity).

Since ε > 0 is arbitrary, |R| < µ?(R) and hence |R| = µ?(R), for any cell R. Example 2.8. The outer measure of Rn is infinite, µ?(Rn) = +∞. For any M > 0, every cell R of volume M is a subset of Rn. Hence, by monotonicity of µ?, µ?(R) ≤ µ?(Rn). But µ?(R) = |R| = M. Thus, µ?(Rn) ≥ M for all M > 0. Thus, µ?(Rn) = +∞. Exercise 13. Show that R is uncountable. Also show that the outer measure of the set of irrationals in R is +∞. Example 2.9. Does the outer measure of other basic subsets, such as balls (or spheres), polygons etc. coincide with their volume, which we know from (calculus)? We shall postpone answering this, in a simple way, till we develop sufficient tools. However, we shall note the fact that for any non-empty open set Ω ⊂ Rn, its outer measure is non-zero, i.e., µ?(Ω) > 0. This is because for every non-empty open set Ω, one can always find a cell R ⊂ Ω such that |R| > 0. Thus, µ?(Ω) ≥ |R| > 0. Exercise 14. If E ⊂ Rn has a positive outer measure, µ?(E) > 0, does there always exist a cell R ⊂ E such that |R| > 0. N CHAPTER 2. LEBESGUE MEASURE ON R 17

Exercise 15. Let E ⊂ Rn and T : Rn → Rn be a linear transformation. Then µ?(T (E)) = |det(T )|µ?(E). Consequently, µ? has the following properties:

(i)( Reflection) µ?(E) = µ?(−E) where −E := {−x | x ∈ E}.

(ii)( Dilation) for λ > 0, µ?(λE) = λnµ?(E) where λE := {λx | x ∈ E}.

(iii) µ? is invariant under rotations.

Recall the definition of outer measure, which was infimum over all covers made up of cells. By a cell, we meant a rectangle in Rn whose sides are parallel to the coordinate axes. As a consequence of above exercise, it turns out that the Lebesgue outer measure is invariant if we include rectangles whose sides are not parallel to the coordinate axes.

Theorem 2.2.4 (Outer Regularity). If E ⊂ Rn is a subset of Rn, then ? ? µ (E) = infΩ⊃E µ (Ω), where Ω’s are open sets containing E. Proof. By monotonicity, µ?(E) ≤ µ?(Ω) for all open sets Ω containing E. Thus, µ?(E) ≤ inf µ?(Ω). Conversely, for each ε > 0, we can choose a cover of cells {Ri} of E such that

∞ X ε |R | ≤ µ?(E) + . i 2 i=1

For each Ri, choose an open cell Qi ⊃ Ri such that ε |Q | ≤ |R | + . i i 2i+1

∞ Since each Qi is open and countable union of open sets is open, Ω = ∪i=1Qi is open. Therefore, by sub-additivity,

∞ ∞ X X ε µ?(Ω) ≤ |Q | ≤ |R | + ≤ µ?(E) + ε. i i 2 i=1 i=1 Thus, we have equality.

Recall that arbitrary union (intersection) of open (closed) sets is open (closed) and finite intersection (union) of open (closed) sets is open (closed). n This motivates us to define the notion of Gδ and Fσ subset of R . N CHAPTER 2. LEBESGUE MEASURE ON R 18

2 Definition 2.2.5. A subset E is said to be Gδ if it is a countable intersection n 3 of open sets in R . We say E is Fσ if it is a countable union of closed sets in Rn.

n n Corollary 2.2.6. For every subset E ⊂ R there exists a Gδ subset G of R such that G ⊃ E and µ?(E) = µ?(G).

Proof. Using Theorem 2.2.4, we have that for every k ∈ N, there is an open set Ωk ⊃ E such that 1 µ?(Ω ) ≤ µ?(E) + . k k ∞ Let G := ∩k=1Ωk. Thus, G is a Gδ set. G is non-empty because E ⊂ G and ? ? hence µ (E) ≤ µ (G). For the reverse inequality, we note that G ⊂ Ωk, for all k, and by monotonicity

1 µ?(G) ≤ µ?(Ω ) ≤ µ?(E) + ∀k. k k Thus, µ?(G) = µ?(E).

Exercise 16 (Continuity from below for outer measure). Let E1,E2,... be n ∞ ? subsets of R such that E1 ⊆ E2 ⊆ ... and E = ∪i=1Ei, then µ (E) = ? limk→∞ µ (Ek).

? ? Proof. By monotonicity of outer measure, we get limk→∞ µ (Ek) ≤ µ (E). It only remains to prove the reverse inequality. Let Gk be a Gδ set such ? that Gk ⊃ Ek, for all k, and µ (Ek) = µ(Gk). The {Gk} may not be an increasing sequence. Hence, we set Fk := ∩j≥kGj and {Fk} is an increasing sequence of measurable sets. Also, Ek ⊂ Fk, since Ek ⊂ Gj for all j ≥ k. ? Moreover, since Ek ⊂ Fk ⊂ Gk, we have µ (Ek) ≤ µ(Fk) ≤ µ(Gk) and hence ? ∞ µ (Ek) = µ(Fk) = µ(Gk). Since E ⊂ ∪k=1Fk

? ∞ ? µ (E) ≤ µ(∪k=1Fk) = lim µ(Fk) = lim µ (Ek). k→∞ k→∞

2terminology comes from German word “Gebiete” and “Durschnitt” meaning territory and average or mean, respectively 3terminology comes from French word “ferm´e”and “somme” meaning closed and sum, respectively N CHAPTER 2. LEBESGUE MEASURE ON R 19

We have checked all the desired properties of µ?, the outer measure, except the countable additivity. If µ? satisfies countable additivity, then we are done with our search for a ‘measure’ generalising the notion of volume. However, unfortunately, it turns out that µ? is not countably additive. In fact, in retrospect, this was the reason for naming µ? as “outer measure” instead of calling it “measure”. Before we show that µ? is not countably additive, we show a property close to additivity.

Proposition 2.2.7. If E and F are subsets of Rn such that d(E,F ) > 0, then µ?(E ∪ F ) = µ?(E) + µ?(F ).

Proof. By the countable sub-additivity of µ?, we have µ?(E ∪ F ) ≤ µ?(E) + µ?(F ). Once we show the reverse inequality, we are done. For each ε > 0, we can choose a covering {Ri} of E ∪ F such that

∞ X ? |Ri| ≤ µ (E ∪ F ) + ε. i=1

The family of cells {Ri} can be categorised in to three groups: those inter- secting only E, those intersecting only F and those intersecting both E and F . Note that the third , cells intersecting both E and F , should have diameter bigger than d(E,F ). Thus, by subdividing these cells to have diameter less than d(E,F ), we can have the family of open cover to consist of only those cells which either intersect with E or F . Let I1 = {i : Ri ∩E 6= ∅} and I2 = {i : Ri ∩ F 6= ∅}. Due to our subdivision, we have I1 ∩ I2 = ∅. Thus, {Ri} for i ∈ I1 is an open cover for E and {Ri} for i ∈ I2 is an open cover for F . Thus,

∞ ? ? X X X ? µ (E) + µ (F ) ≤ |Ri| + |Ri| ≤ |Ri| ≤ µ (E ∪ F ) + ε.

i∈I1 i∈I2 i=1

By the arbitrariness of ε, we have the reverse inequality.

Proposition 2.2.8. If a subset E ⊂ Rn is a countable union of almost ∞ disjoint closed cells, i.e., E = ∪i=1Ri then

∞ ? X µ (E) = |Ri|. i=1 N CHAPTER 2. LEBESGUE MEASURE ON R 20

? P∞ Proof. By countable sub-additivity, we already have µ (E) ≤ i=1 |Ri|. It only remains to prove the reverse inequality. For each ε > 0, let Qi ⊂ Ri be i a cell such that |Ri| ≤ |Qi| + ε/2 . By construction, the cells Qi are pairwise disjoint and d(Qi,Qj) > 0, for all i 6= j. Applying Proposition 2.2.7 finite number times, we have

k ? k  X µ ∪i=1Qi = |Qi| for each k ∈ N. i=1

k Since ∪i=1Qi ⊂ E, by monotonicity, we have

k k ? ? k  X X i µ (E) ≥ µ ∪i=1Qi = |Qi| ≥ |Ri| − ε/2 i=1 i=1 By letting k → ∞, we deduce

∞ X ? |Ri| ≤ µ (E) + ε. i=1 Since ε can be made arbitrarily small, we have equality. A consequence above proposition and Theorem 2.1.4 is that for an open set Ω, ∞ ? X µ (Ω) = |Ri| i=1 irrespective of the choice of the almost disjoint closed cells whose union is Ω. We now show that µ? is not countably additive. In fact, it is not even finitely additive. One should observe that finite additivity of µ? is quite dif- ferent from the result proved in Proposition 2.2.7. To show the non-additivity (finite) of µ?, we need to find two , sum of whose outer measure is not the same as the outer measure of their union4.

∞ Proposition 2.2.9. There exists a countable family {Ni}1 of disjoint sub- sets of Rn such that ∞ ? ∞ X ? µ (∪i=1Ni) 6= µ (Ni). i=1 4This construction is related to the Banach-Tarski paradox which states that one can partition the unit ball in R3 in to a finite number of pieces which can be reassembled (after rotation and translation) to form two complete unit balls! N CHAPTER 2. LEBESGUE MEASURE ON R 21

Proof. Consider the unit cube [0, 1]n in Rn. We define an equivalence relation ∼ (cf. Appendix ??) on [0, 1]n as, x ∼ y whenever x−y ∈ Qn, i.e. we consider the quotient space [0, 1]n/Qn. This equivalence relation will partition the n n cube [0, 1] in to disjoint equivalence classes Eα, [0, 1] = ∪αEα. Now, let N be the subset of [0, 1]n which is formed by picking5 exactly one element from n ∞ each equivalence Eα. Since Q is countable, let {ri}1 be the enumeration n of all elements of Q . Let Ni := N + ri. We first show that Ni’s are all pairwise disjoint. Suppose Ni ∩ Nj 6= ∅, then there exist xα, xβ ∈ N such n that xα + ri = xβ + rj. Hence xα − xβ = rj − ri ∈ Q . This implies that xα ∼ xβ which contradicts that fact that N contains exactly one representative from each equivalence class. Thus, Ni’s are all disjoint. We now show that ∞ n ∞ n ∪i=1Ni = R . It is obvious that ∪i=1Ni ⊂ R . To show the reverse inclusion, n n n we consider x ∈ R . Then there is a rk ∈ Q such that x ∈ [0, 1] + rk. n Hence x − rk ∈ [0, 1] . Thus x belongs to some equivalence class, i.e., there ∞ n is a xα ∈ N such that x − rk ∼ xα. Therefore, x ∈ Nk. Hence ∪i=1Ni = R . Using the sub-additivity and translation-invariance of µ?, we have

∞ ∞ ? n X ? X ? +∞ = µ (R ) ≤ µ (Ni) = µ (N). i=1 i=1 Thus, µ?(N) 6= 0, i.e., µ?(N) > 0. n n Let J := {i | ri ∈ Q ∩ [0, 1] }. Note that J is countable. Now, consider the sub-collection {Nj}j∈J and let F = ∪j∈J Nj. Being a sub-collection of n Ni’s, Nj’s are disjoint. By construction, each Nj ⊆ [0, 2] , and hence F ⊂ [0, 2]n. By monotonicity and volume of cell, µ?(F ) ≤ 2n. If countable- additivity holds true for the sub-collection Nj, then

? X ? X ? µ (F ) = µ (Nj) = µ (N). j∈J j∈J µ?(F ) is either zero or +∞. But we have already shown that µ?(N) 6= 0 and hence µ?(F ) 6= 0. Thus, µ?(F ) = +∞, which contradicts µ?(F ) ≤ 2n. Thus, countable-additivity for Nj cannot hold true. Remark 2.2.10. The clever construction of the set N in the above proof is due to . Thus, the set constructed above is called the . There are more than one Vitali set. Each choice of representative from the equivalence class yields a different Vitali set. 5Possible due to of Choice N CHAPTER 2. LEBESGUE MEASURE ON R 22

k Corollary 2.2.11. There exists a finite family {Ni}1 of disjoint subsets of Rn such that k ? k  X ? µ ∪i=1Ni 6= µ (Ni). i=1 Proof. The proof is ditto till proving the fact that µ?(N) > 0. Now, it is always possible to choose a k ∈ N such that kµ?(N) > 2n. Then, we pick exactly k elements from the set J and for F to be the finite (k) union of k {Nj}1. Now, arguing as above assuming finite-additivity will contradict the fact that kµ?(N) > 2n. The fact that the outer measure µ? is not countably (even finitely) addi- tive6 is a bad news and leaves our job of generalising the notion of volume, for all subsets of Rn, incomplete.

2.2.1 Abstract Set-up Let X be any non-empty set. A set function µ? : 2X → [0, ∞] is called a outer measure on X if

(i) µ?(∅) = 0

∞ ? P∞ ? (ii) If E ⊂ ∪i=1Ei then µ (E) ≤ i=1 µ (Ei). An outer measure µ? is said to be finite if µ?(X) < ∞. An outer measure µ? ? is said to be σ-finite, if there exists subsets {Ei} of X such that µ (Ei) < +∞, ∞ for all i, and X = ∪i=1Ei. ? Exercise 17. Show that there exists subsets {Ei} such that µ (Ei) < +∞ and n ∞ n R = ∪i=1Ei, i.e. R is σ-finite with respect to the Lebesgue outer measure on Rn.

2.3 Measurable Sets

The countable non-additivity of the outer measure, µ?, has left our job in- complete. The next possible attempt is to consider only those subsets of Rn for which µ? is countably additive, i.e. we no longer work in the power set

n 6However, it is possible to get a finitely additive set function on 2R which coincides with µ? on L(Rn) N CHAPTER 2. LEBESGUE MEASURE ON R 23

n n 2R but relax ourselves to a subclass of 2R for which countable additivity holds. For the chosen sub-class of subsets, the notion of volume is generalised and hence we shall call the sub-class ‘measurable’ sets and the outer measure µ? restricted to the sub-class is called the ‘measure’ of the set. However, the difficulty lies in identifying the sub-class? In order to char- acterize the class of measurable sets we observe as a consequence of The- orem 2.2.4 that, for any subset E ⊂ Rn and for each ε > 0, there is an open set Ω ⊃ E such that µ?(Ω) ≤ µ?(E) + ε or µ?(Ω) − µ?(E) ≤ ε. Since Ω = E ∪ Ω \ E is a disjoint union, by demanding countable additivity, we expect to have µ?(Ω) − µ?(E) ≥ µ?(Ω \ E)7. Thus, we need to choose those subsets of Rn for which µ?(Ω) − µ?(E) ≥ µ?(Ω \ E). Definition 2.3.1. We say a subset E ⊂ Rn is measurable (Lebesgue), if for any ε > 0 there exists an open set Ω ⊃ E, containing E, such that µ?(Ω \ E) ≤ ε. Further, we define the measure (Lebesgue), µ, of E as µ(E) = µ?(E). Let L(Rn) denote the class of all subsets of Rn which are Lebesgue mea- n n surable. Thus, L(Rn) ⊂ 2R . The domain of outer measure µ?, is 2R , whereas the domain for the Lebesgue measure, µ, is L(Rn). The inclusion n L(Rn) ⊂ 2R is proper (cf. Exercise 22). By definition, the Lebesgue measure, µ will inherit all the properties of the outer measure µ?. We shall see some examples of Lebesgue measurable subsets of Rn. Example 2.10. It is easy to see that every open set in Rn belongs to L(Rn). Thus, ∅, Rn, open cells etc. are all in L(Rn). Example 2.11. Every subset E of Rn such that µ?(E) = 0 is in L(Rn). By Theorem 2.2.4, for any ε > 0, there is an open set Ω ⊇ E such that µ?(Ω) ≤ µ?(E) + ε = ε. Since Ω \ E ⊆ Ω, by monotonicity of µ?, we have µ?(Ω \ E) ≤ ε. Thus, E ∈ L(Rn). As a consequence, all singletons, finite set, countable sets, Cantor set C in R, Rn−1 ⊂ Rn etc. are all in L(Rn). ∞ n ∞ Theorem 2.3.2. If {Ei}1 is a countable family in L(R ), then E := ∪i=1Ei is in L(Rn), i.e., countable union of measurable sets is measurable.

Proof. Since each Ei is measurable, for any ε > 0, there is an open set Ω ⊃ E such that i i ε µ?(Ω \ E ) ≤ . i i 2i 7The other inequality being true due to sub-additivity N CHAPTER 2. LEBESGUE MEASURE ON R 24

∞ ∞ Let Ω = ∪i=1Ωi, then E ⊂ Ω and Ω is open. But Ω \ E ⊂ ∪i=1(Ωi \ Ei). By monotonicity and sub-additivity of µ?, ∞ ? ? ∞ X ? µ (Ω \ E) ≤ µ (∪i=1Ωi \ Ei) ≤ µ (Ωi \ Ei) ≤ ε. i=1 Thus, E is measurable. Definition 2.3.3. We say E ⊂ Rn is bounded if there is a cell R ⊂ Rn of finite volume such that E ⊂ R. Exercise 18. If E is bounded then show that µ?(E) < +∞. Also, give an example of a set with µ?(E) < +∞ but E is unbounded. Proposition 2.3.4. Compact subsets of Rn are measurable. Proof. Let F be a compact subset of Rn. Thus, µ?(F ) < +∞. By Theo- rem 2.2.4, we have, for each ε > 0, an open subset Ω ⊃ F such that µ?(Ω) ≤ µ?(F ) + ε. If we show µ?(Ω \ F ) ≤ ε, we are done. Observe that Ω \ F is an open set (since F is closed) and hence, by Theorem 2.1.4, there exists almost disjoint closed cells Ri such that ∞ Ω \ F = ∪i=1Ri. k For a fixed k ∈ N, consider the finite union of the closed cells K := ∪i=1Ri. Note that K is compact. Also K ∩F = ∅ and thus, by Lemma ??, d(F,K) > 0. But K ∪ F ⊂ Ω. Thus, µ?(Ω) ≥ µ?(K ∪ F ) (Monotonicity) = µ?(K) + µ?(F ) (By Proposition 2.2.7) ? k ? = µ (∪i=1Ri) + µ (F ) k X ? = |Ri| + µ (F ) (By Proposition 2.2.8). i=1 Pk ? ? Hence, i=1 |Ri| ≤ µ (Ω) − µ (F ) ≤ ε. Since this is true for every k ∈ N, by P∞ taking limit we have i=1 |Ri| ≤ ε. Using Proposition 2.2.8 again, we have ∞ ? X µ (Ω \ F ) = |Ri| ≤ ε. i=1 Thus, F is measurable. N CHAPTER 2. LEBESGUE MEASURE ON R 25

Corollary 2.3.5. Closed subsets of Rn are measurable. Consequently, any Fσ set is measurable.

Proof. Let Γ be a closed subset of Rn. If Γ were bounded, we know it is compact and hence measurable from the above proposition. We need to check only for unbounded set Γ. Let Fi = Γ ∩ Bi(0), for i = 1, 2,.... Note ∞ that each Fi is compact and Γ = ∪i=1Fi. Since each Fi is measurable, using Theorem 2.3.2, we deduce that Γ is measurable. Any Fσ set is a countable union of closed sets and hence is measurable.

Theorem 2.3.6. If E ∈ L(Rn) then Ec ∈ L(Rn).

n Proof. Since E ∈ L(R ), for each k ∈ N, there exists an open set Ωk ⊃ E such ? c n ∞ c that µ (Ωk \ E) ≤ 1/k. Since Ωk is closed, it is in L(R ). Set F := ∪k=1Ωk. c c Note that F is a Fσ set and hence measurable. Since Ωk ⊂ E for every k, c c we have F ⊂ E . Also E \ F ⊂ Ωk \ E, for all k ∈ N. By monotonicity, µ?(Ec \ F ) ≤ 1/k, for all k ∈ N. Therefore, µ?(Ec \ F ) = 0 and hence is measurable. Now Ec = (Ec \ F ) ∪ F is a union of two measurable sets and hence is measurable.

∞ n ∞ Corollary 2.3.7. If {Ei}1 is a countable family in L(R ), then E := ∩i=1Ei is in L(Rn), i.e., countable intersection of measurable sets is measurable. Consequently, any Gδ set is measurable.

∞ ∞ c c Proof. Note that E = ∩i=1Ei = (∪i=1Ei ) .

Exercise 19. If E,F ∈ L(Rn) then show that E \ F ∈ L(Rn).

Theorem 2.3.8 (Borel Regularity). For any subset E ⊂ Rn, the following are equivalent:

(i) E ∈ L(Rn). (ii) For each ε > 0, there is an open set Ω ⊃ E such that µ(Ω \ E) ≤ ε.

(iii)( Inner regularity) For each ε > 0, there is a Γ ⊂ E such that µ(E \ Γ) ≤ ε.

n ? (iv) There exists a Fσ subset F of R such that F ⊂ E and µ (E \ F ) = 0. N CHAPTER 2. LEBESGUE MEASURE ON R 26

Proof. (i) implies (ii). Let E be measurable. Thus, for each ε > 0, there is an open set Ω ⊃ E such that

µ?(Ω \ E) ≤ ε.

Since Ω \ E = Ω ∩ Ec, it is measurable (intersection of measurable sets). Thus, µ?(Ω \ E) = µ(Ω \ E) ≤ ε. (i) and (ii) implies (iii). Let E be measurable and, thus, Ec is measurable. Applying (ii) to Ec we have, for each ε > 0, there is an open set Ω ⊃ Ec such that µ?(Ω \ Ec) ≤ ε. Set Γ := Ωc. Then Γ ⊂ E. Note that E \ Γ = Ω \ Ec. Hence µ(E \ Γ) ≤ ε. (iii) implies (iv). Using (iii), we have that for every k ∈ N, there is a closed set Γk ⊂ E such that 1 µ(E \ Γ ) ≤ . k k

∞ Let F := ∪k=1Γk. Thus, F is a Fσ set and F ⊂ E. Note that E \F ⊂ E \Γk, for each k. Hence, by monotonicity, µ(E \ F ) ≤ µ(E \ Γk) ≤ 1/k for all k. Thus, µ(E \ F ) = 0. (iv) implies (i). Assume (iv). Since F is a Fσ set, it is measurable. And since E \ F has outer measure zero, we have E \ F is measurable. Now, since E = F ∪ (E \ F ), it is measurable.

Exercise 20. If E ⊂ Rm and F ⊂ Rn are measurable subsets, then E × F ⊂ Rm+n is measurable and µ(E ×F ) = µ(E)µ(F ). (Hint: do by cases, for open sets, Gδ sets, measure zero sets and then arbitrary sets.) We have reached the climax of our search for a notion that generalises volume of cells. We shall now show that for the collection of Lebesgue measur- able sets L(Rn), countable additivity is true. We proved in Proposition 2.2.7 and 2.2.8 special cases of countable additivity.

∞ Theorem 2.3.9 (Countable Additivity). If {Ei}1 are collection of disjoint ∞ measurable sets and E := ∪i=1Ei, then

∞ X µ(E) = µ(Ei). i=1 N CHAPTER 2. LEBESGUE MEASURE ON R 27

Proof. Note that E is measurable, since it is countable union of measurable sets. Therefore, by sub-additivity, we have

∞ X µ(E) ≤ µ(Ei). i=1

We need to prove the reverse inequality. Let us first assume that each Ei is bounded. Then, by inner regularity, there is a closed set Fi ⊂ Ei such that ? i ? ? i µ (Ei \ Fi) ≤ ε/2 . By sub-additivity, µ (Ei) ≤ µ (Fi) + ε/2 . Each Fi is also pairwise disjoint, bounded and hence compact. Thus, by Lemma ??, d(Fi,Fj) > 0 for all i 6= j. Therefore, using Proposition 2.2.7, we have for every k ∈ N, k k  X µ ∪i=1Fi = µ(Fi). i=1 k Since ∪i=1Fi ⊂ E, by monotonicity, we have

k k X X  ε  µ(E) ≥ µ ∪k F  = µ(F ) ≥ µ(E ) − . i=1 i i i 2i i=1 i=1 Since the above inequality is true for every k and arbitrarily small ε, we get P∞ µ(E) ≥ i=1 µ(Ei) and hence equality. Let Ei be unbounded for some or all i. Consider a sequence of cells ∞ ∞ n {Rj}1 such that Rj ⊂ Rj+1, for all j = 1, 2,..., and ∪j=1Rj = R . Set Q1 := R1 and Qj := Rj \ Rj−1 for all j ≥ 2. Consider the measurable subsets Ei, j := Ei ∩ Qj. Note the each Ei,j is pairwise disjoint and are each ∞ bounded. Observe that Ei = ∪j=1Ei,j and is a disjoint union. Therefore, P∞ µ(Ei) = j=1 µ(Ei,j). Also, E = ∪i ∪j Ei,j and is a disjoint union. Hence,

∞ ∞ ∞ X X X µ(E) = µ(Ei,j) = µ(Ei). i=1 j=1 i=1

Exercise 21. If E ⊆ F and µ(E) < +∞, then show that µ(F \ E) = µ(F ) − µ(E).

n Corollary 2.3.10. Let E1,E2,... be measurable subsets of R . N CHAPTER 2. LEBESGUE MEASURE ON R 28

∞ (i)( Continuity from below) If E1 ⊆ E2 ⊆ ... and E = ∪i=1Ei, then µ(E) = limk→∞ µ(Ek).

(ii)( Continuity from above) If E1 ⊇ E2 ⊇ ..., µ(Ei) < +∞, for some ∞ i, and E = ∩i=1Ei, then µ(E) = limk→∞ µ(Ek).

Proof. (i) Let F1 := E1 and Fi := Ei \ Ei−1, for i ≥ 2. By construc- ∞ tion, Fi are measurable, disjoint and E = ∪i=1Fi. Hence, by countable additivity,

∞ k X X k µ(E) = µ(Fi) = lim µ(Fi) = lim µ(∪i=1Fi) = lim µ(Ek). k→∞ k→∞ k→∞ i=1 i=1

(ii) Without loss of generality, we assume that µ(E1) < +∞. Set Fi = ∞ Ei \ Ei+1, for each i ≥ 1. Note that E1 = E ∪ (∪i=1Fi) is a disjoint union of measurable sets. Hence,

∞ ∞ X X µ(E1) = µ(E) + µ(Fi) = µ(E) + (µ(Ei) − µ(Ei+1) i=1 i=1 k−1 X = µ(E) + lim (µ(Ei) − µ(Ei+1)) k→∞ i=1 = µ(E) + µ(E1) − lim µ(Ek) k→∞ lim µ(Ek) = µ(E). k→∞

Remark 2.3.11. Observe that for continuity from above, the assumption µ(Ei) < +∞ is very crucial. For instance, consider Ei = (i, ∞) ⊂ R. Note that each µ(Ei) = +∞ but µ(E) = 0. Example 2.12. Recall that in Example 2.6 we proved an inequality regarding the generalised Cantor set C. We now have enough tools to show the equality. Note that each measurable Ck’s satisfy the hypothesis of continuity from above and hence C is measurable and

k µ(C) = lim 2 a1a2 . . . ak. k In view of this example and Proposition A.0.6, we have generalised Cantor set whose outer measure is positive. N CHAPTER 2. LEBESGUE MEASURE ON R 29

Recall that we showed the inner regularity of µ in Theorem 2.3.8. We can, in fact, better this for sets with finite measure.

Corollary 2.3.12. If µ(E) < +∞ then there exists a compact set K ⊂ E such that µ(E \ K) ≤ ε.

Proof. We have, using (iii) of Theorem 2.3.8, that a closed set Γ ⊂ E such that µ(E \ Γ) ≤ ε. Let Ki := Γ ∩ Bi(0) be a sequence of compact sets such ∞ that Γ = ∪i=1Ki and K1 ⊂ K2,.... Therefore, E \ K1 ⊃ E \ K2 ⊃ ... and ∞ E \ Γ = ∩i=1(E \ Ki). Using, continuity from above and µ(E) < +∞, we get

ε ≥ µ(E \ Γ) = lim µ(E \ Ki). i→∞

Thus, for i large enough, we have µ(E \ Ki) ≤ ε.

∞ n Theorem 2.3.13 (First Borel-Cantelli Lemma). If {Ei}1 ⊂ L(R ) be a n P∞ countable collection of measurable subsets of R such that i=1 µ(Ei) < ∞. ∞ ∞ Then E := ∩k=1 ∪i=k Ei has measure zero.

∞ ∞ Proof. Let Fk := ∪i=kEi. Note that F1 ⊃ F2 ... and E = ∩k=1Fk. Let P P∞ i µ(Ei) = L. By countable additivity, µ(F1) ≤ i=1 µ(Ei) = L < ∞. By continuity from above, we have

∞ k−1 ∞ X X µ(E) = lim µ(Fk) = lim µ(∪i=kEi) ≤ lim µ(Ei) = lim (L − µ(Ei)). k→∞ k→∞ k→∞ k→∞ i=k i=1

Thus, µ(E) = 0.

The set E in First Borel-Cantelli lemma is precisely the set of all x ∈ Rn n such that x ∈ Ei for infinitely many i. Let x ∈ R be such that x ∈ Ei only for finitely many i. Arrange the indices i in increasing order, for which x ∈ Ei and let K be the maximum of the indices. Then, x∈ / Fj for all j ≥ K + 1 and hence not in E. Conversely, if x∈ / E, then there exists a j j−1 such that x∈ / Fk for all k ≥ j. Thus, either x ∈ ∪i=1 Ei or x∈ / Ei for all i. Exercise 22. Show that the Vitali set N constructed in Proposition 2.2.9 is n not in L(Rn). Thus, L(Rn) ⊂ 2R is a strict inclusion. Exercise 23. Consider the Vitali set N constructed in Proposition 2.2.9. Show that every measurable subset E ⊂ N is of zero measure. N CHAPTER 2. LEBESGUE MEASURE ON R 30

Proof. Let E ⊂ N be a measurable set such that µ(E) > 0, then for each ri ∈ Q ∩ [0, 1], we set Ei := E + ri and Ni := N + ri. Since Ni’s are disjoint, ∞ Ei’s are disjoint and are measurable. Since ∪i=1Ei ⊂ [0, 2], we have ∞ ∞ ∞ X X 2 ≥ µ(∪i=1Ei) = µ(Ei) = µ(E) = +∞. i=1 i=1 A contradiction due to the assumption that µ(E) > 0. Hence µ(E) = 0.

Exercise 24. If E ∈ L(Rn) such that µ(E) > 0 then show that E has a non-measurable subset. Proof. We first show that every measurable set E ⊂ [0, 1] such that µ(E) > 0 has a non-measurable subset. Consider the non-measurable subset N of [0, 1]. For each ri ∈ Q, we set Ni := N + ri and each of them are non-measurable. ∞ Also, we know that R = ∪i=1Ni. Let E ⊂ [0, 1] be a measurable subset such ∞ ∞ that µ(E) > 0. Set Ei := E ∩ Ni. Note that ∪i=1Ei = E ∩ (∪i=1Ni) = E ∩ R = E. If Ei were measurable, for each i, then being a subset of Ni, ? µ (Ei) = 0. Thus, ∞ ? ∞ X ? 0 < µ(E) = µ (∪i=1Ei) = µ (Ei) = 0, i=1 a contradiction. Thus, our assumption that Ei are measurable is incorrect. Thus, Ei’s are non-measurable subsets of E. Now, let E ⊂ R be any measurable subset such that µ(E) > 0. Note that E = ∪i∈Z (E ∩ [i, i + 1)), where [i, i + 1) are disjoint measurable subsets of R. Hence, X 0 < µ(E) = µ(E ∩ [i, i + 1)). i∈Z Thus, for some i, µ(E ∩[i, i+1)) > 0. For this i, set F := E ∩[i, i+1). Then F − i ⊂ [0, 1] which has positive measure and by earlier argument contains a non-measurable set M. Thus, M + i ⊂ F ⊂ E is non-measurable. Exercise 25. Construct an example of a continuous function that maps a measurable (Lebesgue) set to a non-measurable set.

Definition 2.3.14. We say a subset E ⊂ Rn satisfies the Carath´eodory criterion if

? ? ? c n µ (S) = µ (S ∩ E) + µ (S ∩ E ) for all subsets S ⊂ R . N CHAPTER 2. LEBESGUE MEASURE ON R 31

Equivalently,

? ? ? n µ (S) = µ (S ∩ E) + µ (S \ E) for all subsets S ⊂ R . Note that since S = (S ∩E)∪(S ∩Ec), by sub-additivity of µ?, we always have µ?(S) ≤ µ?(S ∩ E) + µ?(S ∩ Ec). Thus, in order to check the Carath´eodory criterion of a set E, it is enough to check the inequality

? ? ? c n ? µ (S) ≥ µ (S ∩ E) + µ (S ∩ E ) ∀S ⊂ R and µ (S) < +∞. Note that , intuitively, Carath´eodory criterion classifies those sets that re- spect additivity. The above criterion was given by Constantin Carath´eodory for charac- terising the measurable sets. Some books also start with Carath´eodory ap- proach as the definition for measurability, since it is equivalent to our notion of measurability. Moreover, this has the advantage over our definition that it is topology independent and is a purely set-theoretic definition and will suit well in the abstract set-up.

Theorem 2.3.15. E ∈ L(Rn) if and only if E satisfies the Carath´eodory criterion.

Proof. Let S ∈ L(Rn). Note that if S ∈ L(Rn), then by countable additivity of µ we have the equality. Thus, it is enough to check the Carath´eodory criterion of E with non-measurable sets S. Let S be any subset of Rn such that µ?(S) < +∞. We need to show that

µ?(S) ≥ µ?(S ∩ E) + µ?(S \ E).

Corresponding to the set S, by Corollary 2.2.6, there is a Gδ set (hence measurable) G ⊃ S such that µ(G) = µ?(S). By the countable additivity,

µ?(S) = µ(G) = µ(G ∩ E) + µ(G \ E) ≥ µ?(S ∩ E) + µ?(S \ E) where the last inequality is due to monotonicity. Hence one way implication is proved. Conversely, let E ⊂ Rn satisfy the Carath´eodory criterion. We need to show E ∈ L(Rn). To avoid working with ∞, we assume E to be such that N CHAPTER 2. LEBESGUE MEASURE ON R 32

µ?(E) < +∞. We know, by outer regularity, that for each ε > 0 there is an open set Ω ⊃ E such that µ?(Ω) ≤ µ?(E) + ε. But, by Carath´eodory criterion, we have

µ?(Ω) = µ?(Ω ∩ E) + µ?(Ω \ E) = µ?(E) + µ?(Ω \ E).

Thus, µ?(Ω \ E) = µ?(Ω) − µ?(E) ≤ ε. Hence, E is measurable. It now only remains to prove for E such that ? µ (E) = +∞. Let Ei = E ∩ Bi(0), for i = 1, 2,.... Note that each ? ∞ µ (Ei) < +∞ is bounded and E = ∪i=1Ei. Since each Ei is measurable, using Theorem 2.3.2, we deduce that E is measurable.

2.3.1 Abstract Set-up Let X be a non-empty set and 2X is the collection of all subsets of X. We say a sub-collection M ⊂ 2X of subsets of X to be a σ- if

(i) ∅ ∈ M.

(ii) If E ∈ M then Ec ∈ M (closure under complementation).

(iii) If {Ei} ⊂ M then ∪iEi ∈ M (closure under countable union). Exercise 26. Show that

n (a)2 R is a σ-algebra.

(b) L(Rn), the class of all Lebesgue measurable subsets of Rn, forms a σ- algebra.

Let M be a σ-algebra. A set function µ : M → [0, ∞] is called a measure on X if

(i) µ(∅) = 0

∞ P∞ (ii) If E ⊂ ∪i=1Ei then µ(E) ≤ i=1 µ(Ei).

∞ P∞ (iii) If E = ∪i=1Ei is a disjoint union then µ(E) = i=1 µ(Ei). The triplet (X, M, µ) is called a . N CHAPTER 2. LEBESGUE MEASURE ON R 33

Exercise 27. (a) Show that if {µk} is a sequence of measures on the same P σ-algebra M then µ = k µk is also a measure on M.

(b) Show that Lebesgue measure is a measure on L(Rn). (c) Show that the cardinality of a set defines a measure on the σ-algebra 2X . This is called the .

(d) Let X be infinite set. Define, for E ⊂ X, ( 0 E is countable µ(E) = +∞ E otherwise.

Show that µ is a measure on 2X .

(e) Fix a x ∈ X. The at x, for E ⊂ X, is defined as ( 1 x ∈ E δx(E) = 0 otherwise.

X Show that δx is a measure on 2 , for each x. Let X be a and let τ(X) denote the collection of all open subsets of X. The smallest σ-algebra containing τ(X) is said to be the Borel σ-algebra. Every element of the Borel σ-algebra is called a . Let B(Rn) denote Borel σ-algebra of Rn. Exercise 28. Show that B(Rn) ⊂ L(Rn). A measure space (X, M, µ) is said to be complete if every subset of a set of measure zero belongs to M. Exercise 29 (Completeness). If E ∈ L(Rn) such that µ(E) = 0 then show that, for every F ⊂ E, F ∈ L(Rn). Exercise 30. The Lebesgue measure restricted to the Borel σ-algebra is not complete and its completion is L(Rn). A measure µ on a σ-algebra M is said to be a measure if µ(X) = 1 and 0 ≤ µ(E) ≤ 1 for all E ∈ M. The triplet (X, M, µ) is called the . Every element of X is called a sample point. Every element E ∈ M is called an event and µ(E) is the probability of the event E. N CHAPTER 2. LEBESGUE MEASURE ON R 34

2.4 Measurable Functions

Recall that our aim was to develop the Lebesgue notion of integration for functions on Rn. To do so, we need to classify those functions for which Lebesgue integration makes sense. We shall restrict our attention to real valued functions on Rn. Let R := R ∪ {−∞, +∞} denote the extended real line.

Definition 2.4.1. We say a function f on Rn is extended real valued if it takes value on the extended real line R. Henceforth, we will confine ourselves to extended real valued functions unless stated otherwise. By a finite-valued function we will mean a function not taking ±∞. Recall that we said Lebesgue integration is based on the idea of parti- tioning the range. As a simple case, consider the characteristic function of a set E ⊂ Rn, ( 1 if x ∈ E χE(x) = 0 if x∈ / E. We expect that Z Z ? χE(x) = = µ (E) = µ(E). Rn E But the last equality makes sense only when E is measurable. Thus, we expect to compute integrals of only those functions whose range when parti- tioned has the pre- as a measurable subset of Rn.

Definition 2.4.2. We say a function f : E ⊂ Rn → R is measurable (Lebesgue) if for all α ∈ R, the set8

f −1([−∞, α)) = {f < α} := {x ∈ E | f(x) < α} is measurable (Lebesgue). We say f is Borel measurable if {f < α} is a Borel set. A complex valued function f : E ⊂ Rn → C is said to be measurable if both its real and imaginary parts are measurable.

Exercise 31. Every finite valued Borel is Lebesgue mea- surable. 8Usually, called as the sublevel set N CHAPTER 2. LEBESGUE MEASURE ON R 35

Proposition 2.4.3. If f is a finite-valued continuous function on Rn then f is Borel measurable. Consequently, every continuous function is Lebesgue measurable.

Proof. Consider the open interval I = (−∞, α) in R. Since f is continuous, f −1(I) = {f < α} is an open subset of Rn and hence is a Borel set. Thus, f is Borel measurable.

The characteristic function χE is measurable but not continuous for a proper measurable subset of Rn. Henceforth, by measurable function, we shall mean Lebesgue measurable function and by Borel function, we shall mean a Borel measurable function.

Exercise 32. (i) E is measurable iff χE is measurable.

(ii) f is measurable iff {f ≤ α} is measurable for every α ∈ R. (iii) f is measurable iff {f > α} is measurable for every α ∈ R. (iv) If f is measurable then show that −f is also measurable. (v) Let f be finite-valued. f is measurable iff {α < f < β} is measurable for every α, β ∈ R. Exercise 33. If f is measurable then (i) f k, is measurable for all k ≥ 1.

(ii) f + λ is measurable for a given constant λ ∈ R. (iii) λf is measurable for a given constant λ ∈ R. Exercise 34. If f, g are measurable and both finite-valued then both f + g and fg are measurable. Definition 2.4.4. A property is said to hold (a.e.) if it holds except possibly on a set of measure zero. Consequently, we say a measurable function is finite a.e. if the set on which it takes ±∞ is of measure zero. All the “finite-valued” statements above can be replaced with “finite a.e.”. Let M(Rn) denote the space of all finite a.e. measurable functions on Rn. The class of functions M(Rn) forms a vector space over R. Note that M(Rn) excludes those measurable function which takes values on extended line on a non-zero measure set. N CHAPTER 2. LEBESGUE MEASURE ON R 36

Definition 2.4.5. We say two functions f, g are equal a.e., f = g a.e., if the set {x | f(x) 6= g(x)}. is of measure zero.

Define the equivalence relation f ∼ g if f = g a.e. on M(Rn). Thus, we have the quotient space M(Rn)/ ∼. However, as an abuse of nota- tion, it has become standard to identify the quotient space M(Rn)/ ∼ with M(Rn). Henceforth, by M(Rn) we refer to the quotient space. Thus, when- ever we say A ⊂ M(Rn), we usually mean the inclusion of the quotient spaces A/ ∼⊂ M(Rn)/ ∼. In other words, each equivalence class of M(Rn) has a representative from A. Note that the support of a function f ∈ M(Rn) is defined as the closure of the set E, E := {x | f(x) 6= 0}.

Thus, even though χQ ∼ 0 are in the same equivalence class and represent the n same element in M(R )/ ∼, the support of χQ is R whereas the support of zero function is empty set. Therefore, whenever we say support of a function f ∈ M(Rn) has some property, we usually mean there is a representative in the equivalence class which satisfies the said properties. Exercise 35 ( space). If f is measurable and f = g a.e., then g is measurable.

Theorem 2.4.6. Let f be finite a.e. on Rn. The following are equivalent: (i) f is measurable.

(ii) f −1(Ω) is a measurable set, for every open set Ω in R. (iii) f −1(Γ) is measurable for every closed set Γ in R. (iv) f −1(B) is measurable for every Borel set B in R. Proof. Without loss of generality we shall assume that f is finite valued on Rn. ∞ (i) implies (ii) Let Ω be an open subset of R. Then Ω = ∪i=1(a1, bi) where (ai, bi) are intervals which are pairwise disjoint. Observe that

−1 ∞ −1 f (Ω) = ∪i=1f (ai, bi) ∞ = ∪i=1 ({f > ai} ∩ {f < bi}) . N CHAPTER 2. LEBESGUE MEASURE ON R 37

Since f is measurable, both {f > ai} and {f < bi} are measurable for all i. Since countable union and intersection of measurable sets are measurable, we have f −1(Ω) is measurable. (ii) implies (iii) f −1(Γ) = (f −1(Γc))c. Since Γc is open, f −1(Γc) is mea- surable and complement of measurable sets are measurable. (iii) implies (iv) Consider the collection

−1 n A := {F ⊂ R | f (F ) ∈ L(R )}.

We note that the collection A forms a σ-algebra. Firstly, ∅ ∈ A. Let F ∈ A. −1 c −1 −1 −1 c −1 ∞ ∞ −1 f (F ) = f (R) \ f (F ) = (f (F )) . Also, f (∪i=1Fi) = ∪i=1f (Fi). By (iii), every closed subset Γ ∈ A and hence the Borel σ-algebra of R is included in A. (iv) implies (i) Let f −1(B) be measurable for every Borel set B ⊂ R. In particular, (−∞, α) is a Borel set of R, for any α ∈ R, and hence {f < α} is measurable. Thus, f is measurable.

Exercise 36. If f : R → R is monotone then f is Borel. The above exercise is useful in giving the example of a set which is Lebesgue measurable, but not Borel. Let g : [0, 1] → [0, 1] be the function defined as

g(y) = inf {fC (x) = y}, (2.4.1) x∈[0,1] where fC is the Cantor function (cf. AppendixA) Exercise 37. Show that g : [0, 1] → C is bijective and increasing. Conse- quently, g is Borel. Example 2.13 (Example of a Measurable set which is not Borel). Let N be the non-measurable (Lebesgue) Vitali subset of [0, 1] (constructed in Propo- sition 2.2.9). Let M := g(N) is a subset of C. Since µ?(C) = 0, we have by monotonicity µ?(M) = 0 and thus M is Lebesgue measurable (zero outer measure sets are measurable). If M were a Borel set then, by Borel measur- ability of g, N = g−1(M) is also Borel, a contradiction. We provide another example using product of measures. Let N be the non-measurable (Lebesgue) Vitali subset of R. Then N ×{0} ⊂ R2 has outer measure zero and hence is measurable subset of R2. If N × {0} were Borel set of R2 then N should be a Borel set of R (since it is a section of N × {0} with y coordinate fixed). But N is not Borel. N CHAPTER 2. LEBESGUE MEASURE ON R 38

What about composition of measurable functions. Let f : Rn → R be measurable (Lebesgue) and g : R → R is also Lebesgue measurable. Then g ◦ f : Rn → R need not be measurable because g−1(−∞, α) need not be a Borel set. However, by relaxing the condition on g, we may expect the composition to be measurable.

Proposition 2.4.7. If f is measurable, finite a.e. on Rn and g is Borel measurable on R then g ◦ f is (Lebesgue) measurable. In particular, g ◦ f is measurable for continuous function g. Consequently, f +, f −, |f| and |f|p for all p > 0 are all measurable. Also, for any two finite valued measurable functions f, g, max(f, g) and min(f, g) are measurable.

Proof. Consider the set the interval (−∞, α) in R. By the Borel measurabil- ity of g, Ω := g−1(−∞, α) is a Borel set of R. Using the measurability of f, (g ◦ f)−1(−∞, α) = f −1(Ω) is measurable. Prove the rest as an exercise.

Example 2.14. The reverse composition f ◦ g, in general, is not (Lebesgue) measurable. Consider the Borel function g : [0, 1] → C given in (2.4.1). Let N be the non-measurable subset of [0, 1]. Set E := g(N). Since E ⊂ C, E is Lebesgue measurable. Now, set f := χE, which is measurable on [0, 1]. Observe that the composition f ◦ g = χN , is not Lebesgue measurable since N is a non-measurable set. Exercise 38. We showed if f is measurable then |f| is measurable. The converse is not true. Given an example of a non-measurable function f such that |f| is measurable.

Proof. Let N denote the non-measurable subset (say, Vitali set) of [0, 1]. Define f : [0, 1] → R as ( 1 x ∈ N f(x) = −1 x ∈ N c

Then f is non-measurable but |f| is the constant function 1 which is mea- surable.

Proposition 2.4.8. If {fi} are a sequence of measurable functions then supi fi(x), infi fi(x), lim supi→∞ fi(x) and lim infi→∞ fi(x) are all measur- able. Consequently, f(x) := lim fi(x), if exists, is measurable. N CHAPTER 2. LEBESGUE MEASURE ON R 39

Proof. If f(x) := supi fi(x) then {f > a} = ∪i{fi > a} and is measurable. If f(x) := infi fi(x) then f(x) = − supi(−fi(x)). Also, lim supi→∞ fi(x) = infj(supi≥j fi) and lim infi→∞ fi(x) = supj(infi≥j fi). The space of measurable functions M(Rn) is closed under point-wise con- vergence. If C(Rn) denotes the space of all continuous functions on Rn, then we have already seen that C(Rn) ⊂ M(Rn). We know from classical analysis that C(Rn) is not closed under point-wise convergence and M(Rn) can be thought of as the “completion” of C(Rn) under point-wise convergence. Recall that in Riemann integration, we approximated the graph of a given function by polygons, equivalently, we were approximating the given function by step functions. We shall now introduce a general class of functions which includes the step functions which corresponds to Lebesgue intergation. Definition 2.4.9. A finite of characteristic functions is called a , i.e., a function φ : E ⊂ Rn → R is said to be a simple function if it is of the form

k X φ(x) = aiχEi i=1 n for measurable subsets Ei ⊂ R with µ(Ei) < +∞ and ai ∈ R, for all i.A simple function φ is said to be a if Ei = Ri are the (bounded) cells in Rn. By definition, a simple function is measurable and finite, hence in M(Rn). The class of a simple functions forms a vector subspace of M(Rn) as seen from the exercise below. Exercise 39. If φ and ψ are simple functions on Rn then φ + ψ and φψ are simple too. Also, if φ is simple, λφ is simple for all λ ∈ R. Note that the representation of the simple function φ, by our definition, is not unique. Definition 2.4.10. A non-zero simple function φ is said to have the canon- ical representation if k X φ(x) = aiχEi i=1 n for disjoint measurable subsets Ei ⊂ R with µ(Ei) < +∞ and ai 6= 0, for all i, and ai 6= aj for i 6= j. N CHAPTER 2. LEBESGUE MEASURE ON R 40

Exercise 40. Every non-zero simple function can be decomposed in to its unique canonical representation. Proof. Let φ be a simple function. Then φ can take only finitely many distinct values. Let {b1, . . . , bk} be the distinct non-zero values attained by n φ. Define Ei := {x ∈ R | φ(x) = bi}. By definition, Ei’s are disjoint and we have the canonical representation of φ. P Exercise 41. If φ is simple with canonical representation φ = aiχE then P i i |φ| is simple and |φ| = i |ai|χEi . Theorem 2.4.11. For any finite a.e. measurable function f on Rn such that ∞ f ≥ 0, there exists a sequence of simple functions {φk}1 such that

(i) φk ≥ 0, for each k, (non-negative)

(ii) φk(x) ≤ φk+1(x) (increasing sequence) and

(iii) limk→∞ φk(x) = f(x) for all x (point-wise convergence). Proof. Note that the domain of the given function may be of infinite measure. We begin by assuming that f is bounded, |f(x)| ≤ M, and the support of f 9 is contained in a cell RM of equal side length M (a cube) centred at origin . We now partition the range of f, [0,M] in the following way: at every stage k ≥ 1, we partition the range [0,M] with intervals of length 1/2k and correspondingly define the set,  i i + 1 E = x ∈ R | ≤ f(x) ≤ for all integers 0 ≤ i < M2k. i,k M 2k 2k

Thus, for every k ≥ 1, we have a disjoint partition of the domain of f, RM , in to {Ei,k}i. Ei,k are all measurable due to the measurability of f. Hence, for each k, we define the simple function X i φ (x) = χ (x), k 2k Ei,k i∈Ik

k where Ik is the set of all integers in [0,M2 ). By definition φk’s are non- negative and φk(x) ≤ f(x) for all x ∈ RM and for all k. In particular, |φk| ≤ M for all k.

9The reason being a simple function is supported on finite measure set N CHAPTER 2. LEBESGUE MEASURE ON R 41

We shall now show that φk’s are an increasing sequence. Fix k and let k x ∈ RM . Then x ∈ Ej,k, for some j ∈ Ik, and φk(x) = j/2 . Similarly, there 0 0 k+1 is a j ∈ Ik+1 and φk+1(x) = j /2 . The way we chose our partition, we 0 know that j = 2j or 2j + 1. Therefore, φk(x) ≤ φk+1(x). Also, by definition of φk, φk(x) ≤ f(x) for all x ∈ RM . It now remains to show the convergence. Now, for each x ∈ RM ,

j + 1 j 1 |f(x) − φk(x)| ≤ − = . 2k 2k 2k Thus, we have the convergence. For any general non-negative measurable function, we construct a se- quence of bounded functions fk, |fk(x)| ≤ k, supported on a set of finite measure. Consider a cell Rk of equal side length k (a cube) centred at the origin. We define the truncation of f at k level on Rk, for each k, as follows:  f(x) if x ∈ R and f(x) ≤ k  k fk(x) = k if x ∈ Rk and f(x) > k 0 elsewhere.

n k→∞ By construction, fk(x) ≤ fk+1(x), for all x ∈ R . Also, fk(x) −→ f(x) converges point-wise, for all x ∈ Rn. To see this fact, fix x ∈ Rn and let ` be the such that x∈ / Rk for all k < `. Thus, fk(x) = 0 for all k = 1, 2, . . . , ` − 1. Let m = f(x). If m ≤ ` then fk(x) = f(x) for all k ≥ ` and hence the sequence converges point-wise. If m > `, choose the first integer i such that ` + i > m > ` and we have fk(x) = f(x) for all k ≥ ` + i, and converges to f(x). Note that each fk is measurable due to the measurability of f. Since the range of fk is [0, k], it is enough to check the measurablility of fk for c all α ∈ [0, k]. The extreme cases, {fk ≤ 0} = Rk, {fk < 0} = ∅ and n c {fk ≤ k} = R are measurable. For any α ∈ (0, k), {fk ≤ α} = Rk ∪{f ≤ α} is measurable. For each k, fk is non-negative bounded measurable function supported on a set of finite measure. Thus, for each k, we have a sequence of simple functions ψk` satisfying the required properties and ψk` → fk, as ` → ∞. We pick the diagonal sequence φk = ψkk. Note that φk is increasing sequence because ψk`(x) ≤ ψ(k+1)`. Also, 1 |φ (x) − f(x)| ≤ |ψ (x) − f (x)| + |f (x) − f(x)| ≤ + |f (x) − f(x)|. k kk k k 2k k N CHAPTER 2. LEBESGUE MEASURE ON R 42

Thus, we have the point-wise convergence for all x. Exercise 42. Show that in the result proved above if, in , f is bounded (f(x) ≤ M) then the convergence is uniform. Exercise 43. Let f be a non-negative measurable function on Rn. Show that n there exists a sequence of measurable subsets {Ek} of R such that ∞ X 1 f = χ . k Ek k=1

n Proof. Let x ∈ R be such that x ∈ {f = 0}. Then we need to define Ek such that ∞ X 1 χ (x) = 0. k Ek k=1

Equivalently, we need to define Ek such that x∈ / Ek for all k. Thus, {f = 0} ∩ Ek = ∅ for every k. This suggests that on Ek, for every k, f is strictly n positive. Also, if x ∈ R is such that x ∈ Ek for all k, then f(x) = +∞. Let n E1 = {x ∈ R | f(x) ≥ 1} and, for k = 2, 3,..., we define

( k−1 ) 1 X 1 E = x ∈ n | f(x) ≥ + χ (x) . k R k i Ei i=1 By construction, ∞ X 1 f(x) ≥ χ . k Ek k=1 This is because at every stage k,

k X 1 f(x) ≥ χ (x) i Ei i=1 Clearly, the equality is true for {f = 0} and {f = +∞}. Now, fix x ∈ Rn such that 0 < f(x) < +∞, then x∈ / Ekm for a subsequence km of k. Consequently,

k −1 1 Xm 1 f(x) < + χ (x). k i Ei m i=1 P∞ 1 Letting km → ∞, we have f(x) < k=1 k χEk (x). Hence, the equality holds. N CHAPTER 2. LEBESGUE MEASURE ON R 43

In the next result, we relax the non-negativity requirement on f.

Theorem 2.4.12. For any finite a.e. measurable function f on Rn there ∞ exists a sequence of simple functions {Φk}1 such that

(i) |Φk(x)| ≤ |Φk+1(x)| and

(ii) limk→∞ Φk(x) = f(x) for all x (point-wise convergence).

In particular, |Φk(x)| ≤ |f(x)| for all x and k. Proof. Any function f can be decomposed in to non-negative functions as + − follows: f = f − f . Corresponding to each we have a sequence of φk and ψk satisfying properties of previous theorem and converges point-wise + − to f and f , respectively. By setting, Φk := φk − ψk we immediately see that Φk(x) → f(x) for all x. Let E1 := {f < 0}, E2 := {f > 0} and + − E3 := {f = 0}. Since both f and f vanishes on E3, φk, ψk vanishes on E3. Thus, Φk = 0 on E3. Similarly, on E1,Φk = −ψk ≤ 0 and on E2,Φk = φk. Therefore, |Φk| = φk + ψk and hence is an increasing sequence. In the above two theorems, we may allow extended real valued measurable function f, provided we allow the point-wise limit to take ±∞. The results above shows the density of simple functions in the space of finite valued functions in M(Rn) under the topology of point-wise convergence.

2.5 Littlewood’s Three Principles

We have, thus far, developed the notion of measurable sets of Rn and measur- able functions on Rn. J. E. Littlewood10 simplified the connection of theory of measures with classical in the following three observations: (i) Every measurable set is “nearly” a finite union of intervals. (ii) Every measurable function is “nearly” continuous. (iii) Every convergent sequence of measurable functions is “nearly” uni- formly convergent. Littlewood had no contribution in the proof of these principles. He sum- marised the connections of measure theory notions with classical analysis.

10In his book on titled Lectures on the Theory of Functions N CHAPTER 2. LEBESGUE MEASURE ON R 44

2.5.1 First Principle

Theorem 2.5.1 (First Principle). If E is a measurable subset of Rn such that µ(E) < +∞ then, for every ε > 0, there exists a finite union of closed cells, say Γ, such that µ(E4Γ) ≤ ε.11

∞ Proof. For every ε > 0, there exists a closed cover of cells {Ri}1 for E ∞ (E ⊂ ∪i=1Ri) such that

∞ X ε |R | ≤ µ(E) + . i 2 i=1

Since µ(E) < +∞, the series converges. Thus, for the given ε > 0 there exists a k ∈ N such that

k ∞ ∞ X X X ε |R | − |R | = |R | < . i i i 2 i=1 i=1 i=k+1

k Set Γ = ∪i=1Ri. Now,

µ(E4Γ) = µ(E \ Γ) + µ(Γ \ E) (by additivity) ∞  ∞ ≤ µ ∪i=k+1Ri + µ (∪i=1Ri \ E) (by monotonicity) ∞  ∞ = µ ∪i=k+1Ri + µ (∪i=1Ri) − µ(E) (by additivity) ∞ ∞ X X ≤ |Ri| + |Ri| − µ(E) (by sub-additivity) i=k+1 i=1 ε ε < + = ε. 2 2

Note that in the one dimension case one can, in fact, find a finite union of open intervals satisfying above condition. At the end of last section, we saw that the sequence of simple functions were dense in M(Rn) under point-wise convergence. Using, Littlewood’s first principle, one can say that the space of step functions is dense in M(Rn) under a.e. point-wise convergence.

11E4Γ = (E ∪ Γ) \ (E ∩ Γ) N CHAPTER 2. LEBESGUE MEASURE ON R 45

Theorem 2.5.2. For any finite a.e. measurable function f on Rn there exists a sequence of step functions {φk} that converge to f(x) point-wise for a.e. x ∈ Rn.

Proof. It is enough to show the claim for any characteristic function f = χE, where E is a measurable subset of finite measure. By Littlewood’s first ` principle, for every integer k > 0, there exist ∪i=1Ri, finite union of closed ` k cells such that µ(E4 ∪i=1 Ri) ≤ 1/2 . It will be necessary to consider cells that are disjoint to make the value of simple function coincide with f in the intersection. Thus, we extend sides of Ri and form new collection of almost ` m disjoint cells Qi such that ∪i=1Ri = ∪i=1Qi. Further still, for each integer k m k ` k > 0, we can pick disjoint cells ⊂ Qi such that Fk := ∪i=1Pi ⊂ ∪i=1Ri ` k and µ((∪i=1Ri) \ Fk) ≤ 1/2 . Define m X φk := χF = χ k k Pi i=1 is a step function and f(x) = φk(x) for all x ∈ E ∩ Fk and 1 µ({x ∈ n | f(x) 6= φ }) = µ(E4F ) ≤ . R k k 2k−1 ∞ ∞ Observe that G := ∩k=1 ∪i=k (E4Fk) = {f 6→ φk} the set of all points where fails. By first Borel-Cantelli (theorem 2.3.13), P µ(G) = 0 because k(E4Fk) is finite. The space of step functions on Rn is dense in M(Rn) endowed with the point-wise a.e. topology.

2.5.2 Third Principle The time is now ripe to state and prove Littlewood’s third principle, a con- sequence of which is that uniform convergence is valid on “large” subset for a point-wise convergence sequence.

Theorem 2.5.3. Suppose {fk} is a sequence of measurable functions defined on a measurable set E with µ(E) < +∞ such that fk(x) → f(x) (point-wise) ε a.e. on E. Then, for any given ε, δ > 0, there is a measurable subset Fδ ⊂ E ε such that µ(E \ Fδ ) < δ and an integer N ∈ N (independent of x) such that, ε for all x ∈ Fδ , |fk(x) − f(x)| < ε ∀k ≥ N. N CHAPTER 2. LEBESGUE MEASURE ON R 46

Proof. We assume without loss of generality that fk(x) → f(x) point-wise, for all x ∈ E. Otherwise we shall restrict ourselves to the subset of E where it holds and its complement in E is of measure zero. For each ε > 0 and x ∈ E, there exists a k ∈ N (possibly depending on x) such that

|fj(x) − f(x)| < ε ∀j ≥ k. Since we want to get the region of uniform convergence, we accumulate all x ∈ E for which the same k holds for a fixed ε. For the fixed ε > 0 and for each k ∈ N, we define the set

ε Ek := {x ∈ E | |fj(x) − f(x)| < ε ∀j ≥ k} .

ε Note that not all of Ek’s are empty, otherwise it will contradict the point- ε wise convergence for all x ∈ E. Due to the measurability of fk and f, Ek ε ε ∞ ε are measurable. Also, note that by definition Ek ⊂ Ek+1 and ∪k=1Ek = E (Exercise!). Thus, by continuity from below, for each δ > 0, there is a kδ ∈ N such that ε ε µ(E) − µ(Ek) = µ(E \ Ek) < δ ∀k ≥ kδ. If Eε 6= ∅, set F ε := Eε and N := k else set F ε to be the first non-empty kδ δ kδ δ δ ε ε set Em for m ≥ kδ and N := m. We have, in particular, µ(E \ Fδ ) < δ and ε for all x ∈ Fδ , |fj(x) − f(x)| < ε ∀j ≥ N.

Corollary 2.5.4. Suppose {fk} is a sequence of measurable functions defined on a measurable set E with µ(E) < +∞ such that fk(x) → f(x) (point-wise) ε a.e. on E. Then, for any given ε, δ > 0, there is a closed subset Γδ ⊂ E such ε that µ(E \ Γδ) < δ and an integer K ∈ N (independent of x) such that, for ε all x ∈ Γδ, |fk(x) − f(x)| < ε ∀k ≥ K.

ε ε Proof. Using above theorem obtain Fδ such that µ(E \ Fδ ) < δ/2. By inner ε ε ε ε regularity, pick a closed set Γδ ⊂ Fδ such that µ(Fδ \ Γδ) < δ/2.

ε Note that in the Theorem and Corollary above, the choice of the set Fδ ε or Γδ may depend on ε. We can, in fact, have a stronger result that one can choose the set independent of ε. N CHAPTER 2. LEBESGUE MEASURE ON R 47

Corollary 2.5.5 (Egorov). Suppose {fk} is a sequence of measurable func- tions defined on a measurable set E with µ(E) < +∞ such that fk(x) → f(x) (point-wise) a.e. on E. Then, for any given δ > 0, there is a measurable subset Fδ ⊂ E such that µ(E \ Fδ) < δ and fk → f uniformly on Fδ.

Proof. From the theorem proved above, for a given δ > 0 and k ∈ N, there is k a measurable subset Fk ⊂ E such that µ(E \ Fk) < δ/2 and for all x ∈ Fk, there is a Nk ∈ N |fj(x) − f(x)| < 1/k ∀j > Nk. ∞ Set Fδ := ∩k=1Fk. Thus,

∞ ∞ X µ(E \ Fδ) = µ(∪k=1(E \ Fk)) ≤ µ(E \ Fk) < δ. k=1

Now, it is easy to check that fk → f uniformly in Fδ. Example 2.15. The finite measure hypothesis on E is necessary in above theorems. Let fk = χ[k,k+1) then fk(x) → 0 point-wise. Choose δ = 1 and let F be any subset of R such that µ(F ) < 1. Because µ(F ) < 1, c F ∩ [k, k + 1) 6= ∅, for all k. We claim that fk cannot converge uniformly c c to 0 on F . For every k ∈ N, there is a xk ∈ F ∩ [k, k + 1) such that |fk(xk) − f(xk)| = |fk(xk)| = 1. The Egorov’s theorem motivates the following notion of convergence in M(Rn). ∞ Definition 2.5.6. Let {fk}1 and f be finite a.e. measurable functions on a n 12 measurable set E ⊆ R . We say fk converges almost uniformly to f on E, if for every δ > 0, there exists measurable subset Fδ ⊂ E such that µ(Fδ) < δ and fk → f uniformly on E \ Fδ. Exercise 44. Show that almost uniform convergence implies point-wise a.e. convergence.

Proof. Let fk → f almost uniformly converge. Then, by definition, for each k ∈ N, there exists a measurable set Fk ⊂ E such that µ(Fk) < 1/k and ∞ fk → f uniformly on E \ Fk. Let F = ∩k=1Fk. Thus, µ(F ) ≤ µ(Fk) < 1/k for all k and hence µ(F ) = 0. For any x ∈ E \F , x ∈ ∪k=1(E \Fk) and hence

12Note that this notion of convergence is much weaker than demanding uniform conver- gence except on zero measure sets. N CHAPTER 2. LEBESGUE MEASURE ON R 48

x ∈ E \ Fk for some k. Therefore, by the uniform convergence of fn → f, we have fn(x) → f(x) point-wise. Thus, fn(x) → f(x) for all x ∈ E \ F and µ(F ) = 0, showing the point-wise a.e. convergence. The converse is not true. Example 2.15 gives an example of a point-wise a.e. converging sequence which do not converge almost uniformly. However, the converse is true for a finite measure set E. The Egorov’s theorem is pre- cisely the converse statement for finite measure set. Thus, on finite measure set we have the following statement: Exercise 45. For finite measure set µ(E) < +∞, a sequence of functions on E converges point-wise a.e. iff it converges almost uniformly. We shall end this section by giving a weaker notion of convergence on M(Rn). ∞ Definition 2.5.7. Let {fk}1 and f be finite a.e. measurable functions on a n measurable set E ⊆ . We say fk converges in measure to f on E, denoted µ R as fk → f, if for every ε > 0,

ε lim µ(Ek) = 0. k→∞ where ε Ek := {x ∈ E | |fk(x) − f(x)| > ε}. Exercise 46. Almost uniform convergence implies convergence in measure.

Proof. Since fk converges almost uniformly to f. For every ε > 0 and k ∈ N, there exists a set Fk (independent of ε) such that µ(Fk) < 1/k and there exists K ∈ N, for all x ∈ E \ Fk, such that

|fj(x) − f(x)| ≤ ε ∀j > K.

ε Therefore Ej ⊂ Fk for infinitely many j > K. Thus, for infinitely many j, µ(Eε) < 1/k. In particular, choose j > k and µ(Eε ) < 1/k. j k jk Example 2.16. Give an example to show that convergence in measure do not imply almost uniform convergence. However, the converse is true upto a subsequence.

∞ Theorem 2.5.8. Let {fk}1 and f be finite a.e. measurable functions on a n µ measurable set E ⊆ R (not necessarily finite). If fk → f then there is a ∞ subsequence {fkl }l=1 such that fkl converges almost uniformly to f. N CHAPTER 2. LEBESGUE MEASURE ON R 49

2.5.3 Second Principle

Recall M(Rn) is decomposed in to equivalence classes under equality a.e. So, it would be a nice situation if for every measurable function there is a continuous function in its equivalence class. In other words, we wish to have for every measurable function f a continuous function g such that f = g a.e. Unfortunately, this is not true. Exercise 47. Given an example of a measurable function f for which there is no continuous function g such that f = g a.e. Littlewood’s second principle is “approximating” a measurable function on finite measure by a continuous function. n Exercise 48. Let χR be a step function on R , where R is a cell with |R| < n +∞. Then, for ε > 0, there exists a subset Eε ⊂ R such that χR restricted n to R \ Eε is continuous and µ(Eε) < ε. Theorem 2.5.9 (Luzin). Let f be measurable finite a.e. on a measurable set E such that µ(E) < +∞. Then, for ε > 0, there exists a closed set Γε ⊂ E 13 such that µ(E \ Γε) < ε and f |Γε is a continuous function .

Proof. We know from Theorem 2.5.2 that step functions are dense in M(Rn) ∞ under point-wise a.e. convergence. Let {φk}k=1 be a sequence of step func- tions that converge to f point-wise a.e. Fix ε > 0. For each k ∈ N, there k exists a measurable subset Ek ⊂ E such that µ(Ek) < (ε/3)(1/2 ) and φk re- stricted to E \ Ek is continuous. By Egorov’s theorem, there is a measurable set Fε ⊂ E such that µ(E \ Fε) < ε/3 and φk → f uniformly in Fε. Note ∞ that φk restricted to Gε := Fε \ ∪k=1Ek is continuous. Therefore, its uniform limit f restricted to Gε is continuous. Also,

∞ µ(E \ Gε) = µ(∪k=1Ek) + µ(E \ Fε) < (2ε)/3.

By inner regularity, pick a closed set Γε ⊂ Gε such that µ(Gε \ Γε) < ε/3.

Now, obviously, f |Γε is continuous and

µ(E \ Γε) = µ(E \ Gε) + µ(Gε \ Γε) < ε.

13 f restricted to Γε is continuous but f as a function on E may not be continuous on points of Γε N CHAPTER 2. LEBESGUE MEASURE ON R 50

Corollary 2.5.10. Let f be measurable finite a.e. on a measurable set E such that µ(E) < +∞. Then, for ε > 0, there exists a continuous function g on E such that µ({x ∈ E | f(x) 6= g(x)} < ε.

Proof. Use Urysohn lemma or Tietze extension theorem to find a continuous function g on E which coincides with f on Γε. Consider the set

{x ∈ E | f(x) 6= g(x)} = E \ Γε.

The measure of the above set is less than ε.

Exercise 49. For any finite a.e. measurable function f on Rn there exists a sequence of continuous functions {fk} that converge to f(x) point-wise for a.e. x ∈ Rn.

2.6 Jordan Content or Measure

We end this chapter with few remarks on the notion of “Jordan content of a set”, developed by Giuseppe Peano and Camille Jordan. This notion is related to Riemann integration in the same way as Lebesgue measure is related to Lebesgue integation. The Jordan content is the finite version of the Lebesgue measure.

Definition 2.6.1. Let E be bounded subset of Rn. We say that a finite family of cells {Ri}i∈I is a finite covering of E iff E ⊆ ∪i∈I Ri, where I is a finite index set.

n Let B ⊂ 2R be the class of all bounded subsets of Rn. The reason for restricting ourselves to B is because every element of B will admit a finite covering.

Definition 2.6.2. For a subset E ∈ B, we define its Jordan outer content J ?(E) as, ? X J (E) := inf |Ri|, E⊆∪i∈I Ri i∈I the infimum being taken over all possible finite coverings of E. N CHAPTER 2. LEBESGUE MEASURE ON R 51

The term “measure” is usually reserved for a countably additive set func- tion, hence we use the term “content”. Otherwise Jordan content could be viewed as finitely additive measure. Some texts refer to it as or Jordan-Peano measure.

Lemma 2.6.3. The Jordan outer content J ? has the following properties:

(a) For every subset E ∈ B, 0 ≤ J ?(E) < +∞.

(b)( Translation Invariance) For every E ∈ B, J ?(E + x) = J ?(E) for all x ∈ Rn.

(c)( Monotone) If E ⊂ F , then J ?(E) ≤ J ?(F ).

(d)( Finite Sub-additivity) For a finite index set I,

? X ? J (∪i∈I Ei) ≤ J (Ei). i∈I

Proof. The proofs are similar to those of outer measure case.

Exercise 50. Show that J ?(R) = |R| for any cell R ⊂ Rn. Consequently, µ?(R) = J ?(R) for every cell R. Exercise 51. If E ⊂ R2 denotes the region below the graph of a bounded function f :[a, b] → R. Show that the Jordan content of E is same as the Riemann upper sum of f. Example 2.17. Jordan outer content of the empty set is zero, µ?(∅) = 0. Every cell is a cover for the empty set. Thus, infimum over the volume of all cells is zero. Example 2.18. The Jordan outer content for a singleton set {x} in Rn is zero. The same argument as for empty set holds except that now the infimum is taken over all cells containing x. Thus, for each ε > 0, one can find a cell ? Rε such that x ∈ Rε and |Rε| ≤ ε. Therefore, µ ({x}) ≤ ε for all ε > 0 and hence µ?({x}) = 0. Example 2.19. The Jordan outer content of a finite subset E of Rn is zero. A finite set E = ∪x∈E{x}, where the union is finite. Thus, by finite sub- additivity, µ?(E) ≤ 0 and hence µ?(E) = 0. N CHAPTER 2. LEBESGUE MEASURE ON R 52

This is precisely where the difference lies between Lebesgue measure and Jordan content. Recall the Q had Lebesgue outer measure zero. But the Jor- dan outer content of Q contained in a bounded cell is positive. For instance, consider E := Q ∩ [0, 1]. We shall show that J ?(E) = 1. Note that there is no finite cover of E which is properly contained in [0, 1], due to the density of Q in [0, 1]. Thus, any finite cover of E also contains [0, 1]. In fact, [0, 1] is itself a finite cover of E. Infimum over all the finite cover is bounded below by 1, due to monotonicity. Thus, J ?(E) = 1. Exercise 52. Show that J ?(E) = J ?(E), where E denotes the closure of E. Proof. Firstly, the result is true when E = R is a cell, since J ?(R) = µ?(R) = µ?(R) = J ?(R). By monotonicity of J ?, J ?(E) ≤ J ?(E). For converse argument, let {Ri} be any finite cover of E, i.e., E ⊂ ∪iRi then E ⊂ ∪iRi. In general, ∪iRi ⊆ ∪iRi. However, since the union is finite we have equality,

∪iRi = ∪iRi.

Therefore, {Ri} is a finite cover of E. Thus,

? X X J (E) ≤ |Ri| = |Ri|. i i

Taking infimum over all finite covers of E, we get J ?(E) ≤ J ?(E). Note that in the proof above the finite cover played a crucial role. A similar result is not true Lebesgue outer measure. For instance, µ?(Q ∩ [0, 1]) = 0 and its closure is [0, 1] whose outer measure in one. More generally, for every k ≥ 0, we have a set E ⊂ Rn such that, µ?(E) = 0 but µ?(∂E) = k, where ∂E is the boundary of E. For instance, consider E := Qn ∩ [0, k1/n]n. Then, ∂E = [0, k1/n]n and µ?(∂E) = k. Exercise 53. Show that the Jordan content of a set is same as the outer measure of its closure, i.e., µ?(E) = J ?(E). Proof. Note that it is enough to show that µ?(E) = J ?(E). Firstly, it follows from defintion that µ?(E) ≤ J ?(E). For the reverse inequality, we consider {Ri} to be an countable cover of E. Since E is closed and bounded, hence compact, there is a finite sub-cover {Qj} of E. Thus,

? X X J (E) ≤ |Qj| ≤ |Ri|. j i N CHAPTER 2. LEBESGUE MEASURE ON R 53

Taking infimum over all countable covers {Ri}, we get

J ?(E) ≤ µ?(E).

Hence the equality holds. To classify the Jordan measurable subsets, we need to identify the class of subsets of B for which finite additivity holds. Note that Q ∩ [0, 1] cannot be Jordan measurable. Since J ?(Q ∩ [0, 1]) = 1. A similar argument also shows that J ?(Qc ∩ [0, 1] = 1. If finite additivity were true then J ?(Q ∩ [0, 1]) + J ?(Qc ∩ [0, 1]) = 2 6= 1 = J ?([0, 1]).

n Theorem 2.6.4. A E ⊂ R is Jordan measurable iff χE is Riemann integrable.

Now do you see why the characteristic function on Q was not Riemann integrable? Precisely because Q was not Jordan measurable. Do you also see how Lebesgue measure fixes this inadequacy? N CHAPTER 2. LEBESGUE MEASURE ON R 54 Chapter 3

Lebesgue Integration

In this chapter, we shall define the integral of a function on Rn, in a progres- sive way, with increasing order of complexity. Before we do so, we shall state some facts about Riemann integrability in measure theoretic language.

Theorem 3.0.1. Let f :[a, b] → R be a bounded function. Then f ∈ R([a, b]) iff f is continuous a.e. on [a, b]

Basically, the result says that a function is Riemann integrable iff its set of discontinuities are of length (measure) zero.

3.1 Simple Functions

Recall from the discussion on simple functions in previous chapter that the representation of simple functions is not unique. Therefore, we defined the canonical representation of a simple function which is unique. We use this canonical representation to define the integral of a simple function.

Definition 3.1.1. Let φ be a non-zero simple function on Rn having the canonical form k X φ(x) = aiχEi i=1

n with disjoint measurable subsets Ei ⊂ R with µ(Ei) < +∞ and ai 6= 0, for all i, and ai 6= aj for i 6= j. We define the Lebesgue integral of a simple

55 CHAPTER 3. LEBESGUE INTEGRATION 56 function on Rn, denoted as

k Z X φ(x) dµ := aiµ(Ei). n R i=1

n R where µ is the Lebesgue measure on R . Henceforth, we shall denote φ dµ R n as φ dx, for Lebesgue measure. Also, we define the integral of φ on E ⊂ R as, Z Z φ(x) dx := φ(x)χE(x) dx. E Rn

Remark 3.1.2. Consider the zero function as the characteristic function χ∅. Then integral of the zero function is defined as µ(∅) = 0.

Note that the integral of a simple function is always finite. Though, we chose to define integral using the canonical representation, it turns out that integral of a simple function is independent of its representation.

Proposition 3.1.3. For any representation of the simple function φ = Pk i=1 aiχEi , we have k Z X φ(x) dx = aiµ(Ei). n R i=1 Pk Proof. Let φ = i=1 aiχEi be a representation of φ such that Ei’s are pairwise disjoint which is not the canonical form, i.e., ai are not necessarily distinct and can be zero for some i. Let {bj} be the distinct non-zero elements of {a1, . . . , ak}, where 1 ≤ j ≤ k. For a fixed j, we define Fj = ∪i∈I Ei, where Pj Ij := {i | ai = bj}. Then Fj’s are pairwise disjoint and µ(Fj) = i∈I µ(Ei). P j Therefore, φ = j bjχFj is a canonical form of φ. Thus,

k Z X X X X φ(x) dx = bjµ(Fj) = bj µ(Ei) = aiµ(Ei). n R j j i∈Ij i=1

We now consider a representation of φ such that Ei are not necessarily Pk disjoint. Let φ = i=1 aiχEi be any general representation of φ, ai ∈ R. k n Given any collection of subsets {Ei}1 of R , there exists a collection of disjoint m k n m−1 subsets {Fj}1 , for m ≤ 2 , such that ∪jFj = R , ∪j=1 Fj = ∪iEi and, for each i, Ei = ∪j∈Ii Fj where Ii := {j | Fj ⊂ Ei} (Exercise!). For each j, we CHAPTER 3. LEBESGUE INTEGRATION 57 define b := P a where I := {i | F ⊂ E }. Thus, φ = Pm b χ , where j i∈Ij i j j i j=1 j Fj Fj are pairwise disjoint. Hence, from first part of the proof,

m m Z X X X φ(x) dx = bjµ(Fj) = aiµ(Fj) n R j=1 j=1 i∈Ij k k X X X = aiµ(Fj) = aiµ(Ei).

i=1 j∈Ii i=1 Hence the integral is independent of the choice of the representation. Exercise 54. Show the following properties of integral of simple functions:

(i) (Linearity) For any two simple functions φ, ψ and α, β ∈ R, Z Z Z (αφ + βψ) dx = α φ dx + β ψ dx. Rn Rn Rn

(ii) (Additivity) For any two disjoint subsets E,F ⊂ Rn with finite measure Z Z Z φ dx = φ dx + φ dx. E∪F E F

(iii) (Monotonicity) If φ ≥ 0 then R φ ≥ 0. Consequently, if φ ≤ ψ, then Z Z φ dx ≤ ψ dx. Rn Rn

(iv) (Triangle Inequality) We know for a simple function φ, |φ| is also simple. Thus, Z Z

φ dx ≤ |φ| dx. Rn Rn (v) If φ = ψ a.e. then R φ = R ψ. Example 3.1. An example of a Lebesgue integrable function which is not

Riemann integral is the following: Consider the characteristic function χQ. We have already seen in Example 1.3 that this is not Riemann integrable. But Z χQ(x) dx = µ(Q) = 0. Rn CHAPTER 3. LEBESGUE INTEGRATION 58

However, χQ = 0 a.e. and zero function is Riemann integrable. Thus, for χQ which is Lebesgue integrable function there is a Riemann integrable function in its equivalence class. Is this always true? Do we always have a Riemann integrable function in the equivalence class of a Lebesgue integrable function. The answer is a “no”. Find an example! Exercise 55. Show that the Riemann integral and Lebesgue integral coincide for step functions.

3.2 Bounded Function With Finite Measure Support

Now that we have defined the notion of integral for a simple function, we intend to extend this notion to other measurable functions. At this juncture, the natural thing is to recall the fact proved in Theorem 2.4.12, which estab- lishes the existence of a sequence of simple functions φk converging point-wise to a given measurable finite a.e. function f. Thus, the natural way of defining the integral of the function f would be Z Z f(x) dx := lim φk(x) dx. k→∞ Rn Rn This definition may not be well-defined. For instance, the limit on the RHS may depend on the choice of the sequence of simple functions φk.

Example 3.2. Let f ≡ 0 be the zero function. By choosing φk = χ(0,1/k) which converges to f point-wise its integral is 1/k which also converging to zero. However, if we choose ψk = kχ(0,1/k) which converges point-wise to f, R but ψk = 1 for all k and hence converges to 1. But zero function is trivially a simple function with Lebesgue integral zero. Note that the situation is very similar to what happens in Riemann’s notion of integration. Therein we demand that the Riemann upper sum and Riemann lower sum coincide, for a function to be Riemann integrable. In Lebesgue’s situation too, we have that the integral of different of simple functions converging to a function f may not coincide. The follow- ing result singles out a case when the limits of integral of simple functions coincide for any choice. Proposition 3.2.1. Let f be a measurable function finite a.e. on a set E of finite measure and let {φk} be a sequence of simple functions supported on E CHAPTER 3. LEBESGUE INTEGRATION 59

and uniformly bounded by M such that φk(x) → f(x) point-wise a.e. on E. R Then L := limk→∞ E φk dx is finite. Further, L is independent of the choice of {φk}, i.e., if f = 0 a.e. then L = 0.

Proof. Since φk(x) → f(x) point-wise a.e. on E and µ(E) < +∞, by Egorov’s theorem, for a given δ > 0, there exists a measurable subset Fδ ⊂ E R such that µ(E \ Fδ) < δ/(4M) and φk → f uniformly on Fδ. Set Ik := E φk. We shall show that {Ik} is a Cauchy sequence in R and hence converges. Consider Z |Ik − Im| ≤ |φk(x) − φm(x)| (triangle inequality) E Z Z = |φk(x) − φm(x)| + |φk(x) − φm(x)| Fδ E\Fδ Z δ < |φk(x) − φm(x)| + Fδ 2 δ δ < µ(F ) + for all k, m > K δ 2µ(E) 2 ≤ δ for all k, m > K (By monotonicity of µ).

Thus, {Ik} is Cauchy sequence and converges to some L. If f = 0, repeating the above argument on Ik Z |Ik| ≤ |φk(x)| (triangle inequality) E Z Z = |φk(x)| + |φk(x)| Fδ E\Fδ Z δ < |φk(x)| + Fδ 4 3δ δ < µ(F ) + for all k > K0 δ 4µ(E) 2 ≤ δ for all k > K0 (By monotonicity of µ), we get L = 0. We know from Theorem 2.4.12 the existence of a sequence of simple func- tions φk converging point-wise to a given measurable finite a.e. function f. If, in addition, we assume f is bounded and supported on a finite measure set E, then the φk satisfy the hypotheses of above Proposition. This motivates us to give the following definition. CHAPTER 3. LEBESGUE INTEGRATION 60

Definition 3.2.2. Let f be bounded measurable function supported on a set E of finite measure. The integral of f is defined as Z Z f(x) dx := lim φk(x) dx, E k→∞ E where {φk} are uniformly bounded simple functions supported on the support of f and converging point-wise to f. Moreover, for any measurable subset F ⊂ E, Z Z f(x) dx := f(x)χF (x) dx. F E Exercise 56. Show that all the properties of integral listed in Exercise 54 is also valid for an integral of a bounded measurable function with support on finite measure. Exercise 57. A consequence of (v) property is that if f = 0 a.e. then R f = 0. The converse is true for non-negative functions. Let f be a bounded measurable function supported on finite measure set. If f ≥ 0 and R f = 0 then f = 0 a.e. The way we defined our integral of a function, the interchange of limit and integral under point-wise convergence comes out as a gift.

1 Theorem 3.2.3 (Bounded Convergence Theorem ). Let fk be a sequence of measurable functions supported on a finite measure set E such that |fk(x)| ≤ M for all k and x ∈ E and fk(x) → f(x) point-wise a.e. on E. Then f is also bounded and supported on E a.e. and Z lim |fk − f| = 0. k→∞ E In particular, Z Z lim fk = f. k→∞ E E

Proof. Since f is a point-wise a.e. limit of fk, |f(x)| ≤ M a.e. on E and has support in E. By Egorov’s theorem, for a given δ > 0, there exists a measurable subset Fδ ⊂ E such that µ(E \ Fδ) < δ/(4M) and fk → f

1This statement is same as the one in Theorem 1.1.6 where we mentioned the proof is not elementary CHAPTER 3. LEBESGUE INTEGRATION 61

uniformly on Fδ. Also, choose K ∈ N, such that |fk(x) − f(x)| < δ/2µ(E) for all k > K. Consider Z Z Z |fk(x) − f(x)| ≤ |fk(x) − f(x)| + |fk(x) − f(x)| E Fδ E\Fδ δ δ < µ(F ) + for all k > K δ 2µ(E) 2 ≤ δ for all k > K (By monotonicity of µ). R Therefore, limk→∞ E |fk − f| = 0 and, by triangle inequality, Z Z lim fk = f. k→∞ E E

It is now time to address the problem of Riemann integration which does not allow us to interchange point-wise limit and integral. We first observe that Riemann integration is same as Lebesgue integration for Riemann in- tegrable functions and thus, by BCT, we have the interchange of limit and integral for Riemann integrable functions, when the limit is also Riemann integrable.

Theorem 3.2.4. If f ∈ R([a, b]) then f is bounded measurable and

Z b Z f(x) dx = f(x) dx, a [a,b] the LHS is in the sense of Riemann and RHS in the sense of Lebesgue.

Proof. Since f ∈ R([a, b]), f is bounded, |f(x)| ≤ M for some M > 0. Also, the support of f, being subset of [a, b], is finite. We need to check that f is measurable. Since f is Riemann integrable there exists two sequences of step functions {φk} and {ψk} such that

φ1 ≤ ... ≤ φk ≤ ... ≤ f ≤ ψk ≤ ... ≤ ψ1 and Z b Z b lim φk = lim ψk = lim f. k a k a k CHAPTER 3. LEBESGUE INTEGRATION 62

Also, |φk| ≤ M and |ψk| ≤ M for all k. Since Riemann integral is same as Lebesgue integral for step functions,

Z b Z Z b Z φk = φk and ψk = ψk. a [a,b] a [a,b]

Let Φ(x) := limk φk(x) and Ψ := limk ψk(x). Thus, Φ ≤ f ≤ Ψ. Being limit of simple functions Φ and Ψ are measurable and by BCT,

Z Z b Z b Z Φ = lim φ = lim ψ = Ψ. [a,b] k a k a [a,b] R Thus, [a,b](Ψ−Φ) = 0. Moreover, since ψk −φk ≥ 0, we must have Ψ−Φ ≥ 0. Thus, by Exercise 57,Ψ − Φ = 0 a.e. and hence Φ = f = Ψ a.e. Hence f is measurable. Thus,

Z Z Z b f(x) dx = lim φk = f(x) dx. [a,b] k [a,b] a

The same statement is not true, in general, for improper Riemann inte- gration (cf. Exercise 62).

3.3 Non-negative Functions

We have already noted that (cf. Example 3.2) for a general measurable func- tion, defining its integral as the limit of the simple functions converging to it, may not be well-defined. However, we know from the proof of Theorem 2.4.11 that any non-negative function f has truncation fk which are each bounded and supported on a set of finite measure, increasing and converge point-wise to f. There could be many other choices of the sequences which satisfy sim- ilar condition. This motivates a definition of integrability for non-negative functions. Definition 3.3.1. Let f be a non-negative (f ≥ 0) measurable function. The integral of f is defined as, Z Z f(x) dx = sup g(x) dx Rn 0≤g≤f Rn CHAPTER 3. LEBESGUE INTEGRATION 63 where g is a bounded measurable function supported on a finite measure set. As usual, Z Z f(x) dx = f(x)χE(x) dx E Rn since fχE ≥ 0 too, if f ≥ 0. Note that the supremum could be infinite and hence the integral could take infinite value. Definition 3.3.2. We say a non-negative function f is Lebesgue integrable if Z f(x) dx < +∞. Rn Exercise 58. Show the following properties of integral for non-negative mea- surable functions:

(i) (Linearity) For any two measurable functions f, g and α, β ∈ R, Z Z Z (αf + βg) dx = α f dx + β g dx. Rn Rn Rn

(ii) (Additivity) For any two disjoint subsets E,F ⊂ Rn with finite measure Z Z Z f dx = f dx + f dx. E∪F E F

(iii) (Monotonicity) If f ≤ g, then Z Z f dx ≤ g dx. Rn Rn In particular, if g is integrable and 0 ≤ f ≤ g, then f is integrable.

(iv) If f = g a.e. then R f = R g.

(v) If f ≥ 0 and R f = 0 then f = 0 a.e.

(vi) If f is integrable then f is finite a.e. Do we have non-negative functions which are not Lebesgue integrable, i.e., for which the supremum is infinite? CHAPTER 3. LEBESGUE INTEGRATION 64

Example 3.3. The function ( 1 for |x| ≤ 1 f(x) = |x| 0 for |x| > 1. is not integrable. Following the definition of the notion of integral of a function, the im- mediate question we have been asking is the interchange of point-wise limit and integral. Thus, for a sequence of non-negative functions {fk} converging point-wise to f is Z Z f = lim fk. k We have already seen in Example 3.2 that this is not true, in general. How- ever, the following result is always true.

2 Lemma 3.3.3 (Fatou). Let {fk} be a sequence of non-negative measurable functions converging point-wise a.e. to f, then Z Z f ≤ lim inf fk. k

Proof. Let 0 ≤ g ≤ f, where g is bounded with support on a set of finite measure E. Let gk(x) = min(g(x), fk(x)), then gk is measurable. Also, gk is bounded by the bound of g and supported on E, since gk ≤ g and g, fk are non-negative for all k. We claim that gk converges to g point-wise a.e. in E. Fix x ∈ E. Then either g(x) = f(x) or g(x) < f(x). Consider the case when g(x) < f(x). For any given ε > 0 there is a K ∈ N such that |fk(x) − f(x)| < ε for all k ≥ K. In particular, this is true for all ε ≤ f(x) − g(x). Thus, g(x) ≤ f(x) − ε < fk(x) for all k ≥ K and hence gk(x) = g(x) for all k ≥ K. On the other hand if g(x) = f(x) then gk(x) is either f(x) or fk(x) and will converge to f(x). Thus, gk(x) → g(x) a.e. R R and by the BCT gk → g. Moreover, gk ≤ fk and, by monotonicity, R R gk ≤ fk and therefore, Z Z Z Z g = lim gk = lim inf gk ≤ lim inf fk. k k k

2Note that we do not demand integrability CHAPTER 3. LEBESGUE INTEGRATION 65

Thus, Z Z Z f = sup g ≤ lim inf fk. k

Exercise 59. Give an example of a situation where the we have strict inequal- ity in Fatou’s lemma. Observe that Fatou’s lemma basically says that the interchange of limit and integral is valid almost half way and what may go wrong is that Z Z lim sup fk > f. k

Corollary 3.3.4. Let {fk} be a sequence of non-negative measurable func- tions converging point-wise a.e. to f and fk(x) ≤ f(x), then Z Z f = lim fk. k R R Proof. By monotonicity, fk ≤ f and hence Z Z Z lim sup fk ≤ f ≤ lim inf fk. k k

The second inequality is due to Fatou’s lemma and hence we have Z Z f = lim fk. k

Corollary 3.3.5 (Monotone Convergence Theorem). Let {fk} be an increas- ing sequence of non-negative measurable functions converging point-wise a.e. to f, i.e., fk(x) ≤ fk+1(x), for all k, then Z Z f = lim fk. k

Proof. Since fk is increasing sequence, we have fk(x) ≤ f(x) and hence we have our result by previous corollary. CHAPTER 3. LEBESGUE INTEGRATION 66

The MCT is not true for a decreasing sequence of functions. Consider fk = χ[k,∞) on R. fk are non-negative and measurable functions on R. The sequence fk converges point-wise to f ≡ 0, since for each fixed x ∈ R, fk(x) = 0 for infinitely many k’s. However, {fk} are decreasing, fk+1(x) ≤ fk(x) for R R all x and k. Moreover, fk = ∞ and f = 0. Thus, Z Z f 6= lim fk. k

Corollary 3.3.6. Let {fk} be a sequence of non-negative measurable func- tions. Then ∞ ! ∞ Z X X Z fk(x) dx = fk(x) dx. k=1 k=1 Pm P∞ Proof. Set gm(x) = k=1 fk(x) and g(x) = k=1 fk(x). gm are measurable and gm(x) ≤ gm+1(x) and gm converges point-wise g. Thus, by MCT, Z Z g = lim gm. m

Thus,

∞ m ∞ Z X Z Z X Z X Z fk(x) = g = lim gm = lim fk(x) = fk(x). m m k=1 k=1 k=1

The highlight of Fatou’s lemma and its corollary is that they all remain true for a measurable function, i.e., we do allow the integrals to take ∞. We end this section by giving a different proof to the First Borel-Cantelli theorem proved in Theorem 2.3.13.

∞ n Theorem 3.3.7 (First Borel-Cantelli Lemma). If {Ei}1 ⊂ L(R ) be a n P∞ countable collection of measurable subsets of R such that i=1 µ(Ei) < ∞. ∞ ∞ Then E := ∩k=1 ∪i=k Ei has measure zero. P Proof. Define fk := χEk and f = fk. Since fk are non-negative, we have from the above corollary that

Z X f = µ(Ek) < +∞. k CHAPTER 3. LEBESGUE INTEGRATION 67

Thus, f ∈ L1(Rn) and hence is finite a.e. Thus, the set F := {x ∈ Rn | f(x) = ∞} has measure zero. We claim that E = F . If x ∈ F , then P k µ(Ek) = ∞ implies that x ∈ Ek, for infinitely many k (a fact observed before) and hence x ∈ E. Conversely, if x ∈ E, then x ∈ Ek for infinitely many k and hence x ∈ F . Thus, µ(E) = 0.

3.4 General Integrable Functions

In this section, we try to extend our notion of integral to all other measurable functions. Recall that any function f can be decomposed in to f = f + − f − and both f +, f − are non-negative. Note that if f is measurable, both f + and f − are measurable. We now give the definition of the integral of measurable functions. Definition 3.4.1. The Lebesgue integral of any measurable real-valued func- tion f on Rn is defined as Z Z Z f(x) dx = f +(x) dx − f −(x) dx. Rn Rn Rn Any measurable function f is said to be Lebesgue integrable if Z |f(x)| dx < +∞. Rn Any measurable function f is said to be locally Lebesgue integrable if Z |f(x)| dx < +∞, K for all compact subsets K ⊂ Rn. Observe that if f is measurable then |f| is measurable. But the converse is not true (cf. Exercise 38). In view of this, one may have a non-measurable function f which is Lebesgue integrable. To avoid this situation, we assume the measurability of f in the definition of Lebesgue integrability of f. Why use |f| in the definition of integrability? For f to be integrable both f + and f − should both be integrable which is true iff |f| is integrable. Exercise 60. Show that the definition of integral of f is independent of its decomposition f = f1 − f2 where fi ≥ 0 for i = 1, 2. CHAPTER 3. LEBESGUE INTEGRATION 68

1 Exercise 61. The function f :[−1, 0) ∪ (0, 1] → R defined as f(x) = x is not Lebesgue integrable although the improper integral (p.v.) exists.

Proof. The principal value intergal exists because

Z −ε 1 Z 1 1 lim dx + dx = 0. ε→0 −1 x ε x

R 0 R 1 But the Lebesgue integrals are −1 1/x = −∞ and 0 1/x = ∞.

sin x Exercise 62. Consider f(x) = x on [0, ∞). Using one can show that f is Riemann integrable (improper) and is equal to π/2. R ∞ + R ∞ − However, f is not Lebesgue integrable since 0 f = 0 f = ∞. 1 1 1 Exercise 63. The function f : (0, 1] → R defined as f(x) = x sin( x ) + cos( x ) is not Lebesgue integrable although the improper integral exists.

Proof. The improper intergal exists because

Z 1  1 1 1  Z 1 d  1  lim sin( ) + cos( ) dx = lim x cos( ) dx = cos 1. ε→0 ε x x x ε→0 ε dx x

Definition 3.4.2. A complex valued measurable function f = u + iv on Rn is said to be integrable if Z Z |f(x)| dx = u2(x) + v2(x)1/2 dx < +∞. Rn Rn and the integral of f is given by Z Z Z f = u + i v.

Exercise 64. Show that a complex-valued function is integrable iff both its real and imaginary parts are integrable. Exercise 65. Show that all the properties of integral listed in Exercise 54 and Exercise 58 is also valid for a general integrable function. CHAPTER 3. LEBESGUE INTEGRATION 69

The space of all real-valued measurable integrable functions on Rn is denoted by L1(Rn). Thus, L1(Rn) ⊂ M(Rn). We will talk more on these spaces in the next section. We introduce this notation early in here only to use them in the statements of our results. We highlight here that the non-negativity hypothesis in all results proved in the previous section (Fatou’s lemma and its corollaries) can be replaced 1 n with a lower bound g ∈ L (R ), since then we work with gk = fk − g ≥ 0. As usual we prove a result concerning the interchange of limit and integral, called the Dominated Convergence Theorem.

Theorem 3.4.3 (Dominated Convergence Theorem). Let {fk} be a sequence of measurable functions converging point-wise a.e. to f. If |fk(x)| ≤ g(x), for all k, such that g ∈ L1(Rn) then f ∈ L1(Rn) and Z Z f = lim fk. k→∞

1 n Proof. Since |fk| ≤ g, |f| ≤ g and since g ∈ L (R ), by monotonicity, 1 n f ∈ L (R ). Note that fk(x) ≤ |fk(x)| ≤ g(x). Hence, g − fk ≥ 0 and converges point-wise a.e. to g − f. By Fatou’s lemma, Z Z (g − f) ≤ lim inf (g − fk).

Therefore, Z Z Z  Z  Z Z g − f ≤ g + lim inf − fk = g − lim sup fk. R R R Thus, lim sup fk ≤ f, since g is finite. Repeating above argument for R R R the non-negative function g + fk, we get f ≤ lim inf fk. Thus, f = R lim fk.

Exercise 66 (Generalised Dominated Convergence Theorem). Let {gk} ⊂ L1(Rn) be a sequence of integrable functions converging point-wise a.e. to 1 n g ∈ L (R ). Let {fk} be a sequence of measurable functions converging 1 n point-wise a.e. to f and |fk(x)| ≤ gk(x). Then f ∈ L (R ) and further if Z Z lim gk = g k then Z Z lim fk = f. k→∞ CHAPTER 3. LEBESGUE INTEGRATION 70

Example 3.4. The Lebesgue dominated convergence theorem is a weaker k statement than demanding uniform convergence. Consider fk(x) = x on [0, 1). fk(x) → 0 point-wise on [0, 1). However, the convergence is not uni- form. But |fk| ≤ 1 and

Z Z 1 1 xk dx = xk dx = [0,1] 0 k + 1 converges to zero.

Example 3.5. We have already seen using ψk in Example 3.2 that the bound by g in the hypothesis of DCT cannot be done away with. In fact, one can modify ψk in that example to have functions whose integrals diverge. For 2 instance, choose fk(x) = kψk = k χ(0,1/k) which point-wise converges to zero R and fk = k which diverges. Example 3.6. The condition that g ∈ L1(Rn) is also crucial. For instance, let fk(x) = 1/kχ[0,k] and |fk(x)| ≤ 1 on R. Note that fk converge uniformly to R 1 zero, fk = 1 do not converge to zero. Why? Because g ≡ 1 is not in L (R). 1 n Corollary 3.4.4. Let {fk} ⊂ L (R ) such that

∞ X Z |fk| dx < +∞. k=1

P∞ 1 n Then k=1 fk(x) ∈ L (R ) and

∞ ! ∞ Z X X Z fk(x) dx = fk(x) dx. k=1 k=1 P∞ Proof. Let g := k=1 |fk|. Since |fk| is a non-negative sequence, by a corol- lary to Fatou’s lemma, we have that

∞ ! ∞ Z Z X X Z g(x) dx = |fk| dx = |fk| dx < +∞. k=1 k=1

Thus, g ∈ L1(Rn). Now, consider

∞ ∞ X X fk(x) ≤ |fk(x)| = g. k=1 k=1 CHAPTER 3. LEBESGUE INTEGRATION 71

P∞ 1 n 1 n Therefore k=1 fk(x) ∈ L (R ), since g ∈ L (R ). Consider the partial sum

m X Fm(x) = fk(x). k=1

Note that |Fm(x)| ≤ g(x) for all k and Fm(x) → f(x) a.e.. Thus, by DCT, R R limm Fm = f. By finite additivity of integals, we have

m ∞ Z Z X Z X Z f = lim Fm = lim fk = fk. m m k=1 k=1 Hence proved.

Note that the BCT (cf. Theorem 3.2.3) had a stronger statement than DCT above. In fact, we can prove a similar statement for DCT.

Exercise 67. Let {fk} be a sequence of measurable functions converging 1 n point-wise a.e. to f. If |fk(x)| ≤ g(x) such that g ∈ L (R ) then Z lim |fk − f| = 0. k→∞

Proof. Note that |fk − f| ≤ 2g and by DCT Z lim |fk − f| = 0. k

The above exercise could also be proved without using DCT and it is good enough proof to highlight here.

Theorem 3.4.5. Let {fk} be a sequence of measurable functions converging 1 n point-wise a.e. to f. If |fk(x)| ≤ g(x) such that g ∈ L (R ) then Z lim |fk − f| = 0. k→∞

In particular, Z Z lim fk = f. k→∞ CHAPTER 3. LEBESGUE INTEGRATION 72

Proof. Note that |f(x)| ≤ g(x), since f is point-wise limit of fk. Let Ek := {x | x ∈ Bk(0) and g(x) ≤ k}. Note that g is non-negative. Set gk(x) := g(x)χEk (x) is measurable, integrable and non-negative. Also, gk(x) ≤ gk+1(x) and gk(x) converges point-wise to g(x). By MCT, we have Z Z lim gk = g. k

Thus, for any given ε > 0, there exists a K ∈ N such that Z ε g < ∀k ≥ K. c 4 Ek

For the K obtained above, fk restricted to EK is uniformly bounded by K. 0 Since fk(x) → f(x) point-wise a.e. on EK , by BCT, there exists a K ∈ N Z ε 0 |fk − f| < k ≥ K . EK 2 We have Z Z Z |fk − f| = |fk − f| + |fk − f| c EK EK Z Z ≤ |fk − f| + 2 g c EK EK ε ε < + 2 = ε ∀k ≥ K0. 2 4 R Hence limk |fk − f| → 0. The idea of the proof above actually suggests the following result.

Proposition 3.4.6. Let f ∈ L1(Rn). Then, for every given ε > 0, (i) There exists a ball B ⊂ Rn of finite measure such that Z |f| < ε. Bc

(ii) (Absolute Continuity) There exists a δ > 0 such that Z |f| < ε whenever µ(E) < δ. E CHAPTER 3. LEBESGUE INTEGRATION 73

Proof. Let g(x) := |f(x)| and hence g ≥ 0.

(i) Let Bk := Bk(0) denote the ball of radius k centred at origin. Set

gk(x) := g(x)χBk (x) is measurable and non-negative. Also, gk(x) ≤ gk+1(x) and gk(x) converges point-wise to g(x). By MCT, we have Z Z lim gk = g. k R Thus, for the given ε > 0, there exists a K ∈ N such that |f| − R |f|χBk < ε for all k ≥ K. Hence, Z Z

ε > (1 − χBk )|f| = |f| ∀k ≥ K. c Bk

(ii) Let Ek := {x | g(x) ≤ k} and gk(x) = g(x)χEk (x). gk is non-negative measurable function and gk(x) ≤ gk+1(x). Again, by MCT, there exists R R n a K ∈ N such that g − gk < ε/2 for all k ≥ K. For any E ∈ L(R ), Z Z Z g = (g − gK ) + gK E E E Z Z ≤ (g − gK ) + gK E Z ≤ (g − gK ) + Kµ(E)

ε Now, choose δ > 0 such that δ < 2K . If µ(E) < δ then Z Z ε ε g ≤ (g − gK ) + Kµ(E) < + K = ε. E 2 2K

The first part of the proposition above suggests that for integrable func- tions, the “integral of the function” vanishes as we approach infinity. How- ever, this is not same as saying the function vanishes point-wise as |x| ap- proaches infinity. Example 3.7. Consider the real-valued function f on R ( x x ∈ f(x) = Z 0 x ∈ Zc. R f = 0 a.e. and Bc f = 0, however limx→∞ f(x) = +∞. CHAPTER 3. LEBESGUE INTEGRATION 74

Can we have a continuous function in the above example? P∞ Example 3.8. Let f = k=1 kχ[k,k+1/k3). Note that

∞ ∞ X Z X 1 kχ k,k+ 1 dx = 2 < +∞. [ k3 ) k k=1 k=1

The integral of f is ∞ Z X f = 1/k2 k=1 and is in L1(Rn). But lim sup f(x) = +∞. x→+∞

In fact, this is true for a continuous function in L1(R) (extend the f contin- uously to R). However, for a uniformly continuous function in L1(R) we will have lim|x|→∞ f(x) = 0. Example 3.9. The integrability assumption, i.e., f ∈ L1(Rn) is crucial the absolute continuity property (ii). Consider f(x) = 1/x in (0, 1). Then for all R δ δ > 0 0 |f| is not necessarily small. than can be large The absolute continuity property of the integral proved in the Proposition above is precisely the continuity of the integral. Exercise 68. Let f ∈ L1([a, b]) and

Z x F (x) = f(t) dt. a

Then F is continuous on [a, b].

Proof. Let x ∈ (a, b). Consider

Z y Z y

|F (x) − F (y)| = f(t) dt ≤ |f(t)| dt. x x

Since f ∈ L1([a, b]), by absolute continuity, for any given ε > 0 there is a δ > 0 such that for all y ∈ E = {y ∈ [a, b] | |x − y| < δ}, we have |F (x) − F (y)| < ε. CHAPTER 3. LEBESGUE INTEGRATION 75

3.5 Order of Integration

Theorem 3.5.1 (Fubini). Let f : Rm × Rn → R be an integrable function. Then

(i) f y : Rm → R is integrable defined as f y(x) := f(x, y), for a.e. y ∈ Rn and f x : Rn → R is integrable defined as f x(y) := f(x, y), for a.e. x ∈ Rm. (ii) y 7→ R f y(x) dx is integrable on n and x 7→ R f x(y) dy is integrable Rm R Rn on Rm. (iii) R R f y(x) dx = R f(x, y) dx dy = R R f x(y) dy. Rn Rm Rm+n Rm Rn Proof. First we observe that it is enough to prove the results for f y and similar arguments are valid for f x. Let F denote all integrable functions on Rm+n satisfying (i), (ii) and (iii). We have to show that every integrable functions belongs to F.

Step 1 We first observe that F is closed under finite linear combinations. If {fi} is a finite collection in F and {Ai} is the collection of zero y c measure sets such that fi is integrable, for all y ∈ Ai , then ∪iAi is of y measure zero and in its complement fi is integrable for all i. Thus, y if f is any finite linear combination of {fi} then f is integrable in the complement of ∪iAi and (ii) and (iii) follows from the linearity of integral.

Step 2 Let {fk} be an increasing (or decreasing) sequence of non-negative functions in F converging pointwise to f and let us assume that f is integrable. We claim f ∈ F. By MCT, we have Z Z f(x, y) dx dy = lim fk(x, y) dx dy. k→∞ Rm+n Rm+n y Let {Ak} be the collection of zero measure sets such that fk is in- c tegrable, for all y ∈ Ak, then ∪kAk is of measure zero and in its y complement fk is integrable for all k. By MCT, Z y gk(y) := fk (x) dx Rm CHAPTER 3. LEBESGUE INTEGRATION 76

is an increasing sequence converging to Z g(y) := f y(x) dx. Rm By MCT, Z Z g(y) dy = lim gk(y) dy. k→∞ Rn Rn But due to the assumption fk ∈ F we know that Z Z gk(y) dy = fk(x, y) dx dy. Rn Rm+n Thus, we obtain Z Z g(y) dy = f(x, y) dx dy. Rn Rm+n Since f is integrable, g is integrable and, hence, g is finite a.e. Thus, f y is integrable for a.e. y and Z Z  Z f y(x) dx dy = f(x, y) dx dy. Rn Rm Rm+n Hence, f ∈ F.

m+n Step 3 We now claim that χE ∈ F where E is a measurable subset of R with finite measure.

(a) Suppose E is bounded open cell. Then E = Em × En where Em m n y and En are cells of R and R . Then χE is integrable for all y because ( y χEm y ∈ En χE = 0 y∈ / En. and Z ( y |Em| y ∈ En χE(x) dx = = |Em|χEn Rm 0 y∈ / En is also integrable. Therefore, Z Z Z y χE(x) dx dy = |Em||En| = |E| = χE dx dy Rn Rm Rm+n

and χE ∈ F. CHAPTER 3. LEBESGUE INTEGRATION 77

(b) Suppose E is a subset of the boundary of some closed cell. Then R y m m+n χE(x, y) dx dy = 0. Also, χ = 0 a.e. in , for a.e. R E R y ∈ Rn, and Z y χE(x) dx = 0. Rm Hence, Z Z  y χE(x) dx dy = 0 Rn Rm

and χE ∈ F. (c) Suppose E is a finite almost disjoint union of closed cells, i.e. N E = ∪ R . Then χ is a linear combination of χ ◦ , interior k=1 k E Rk of Rk, and Γk, a subset of boundary of Rk. Thus, by Step 1, χE ∈ F. ∞ (d) Suppose E is open with finite measure. Then E = ∪k=1Rk is countable almost disjoint union of closed cells. Then fk := Pk i=1 χRi increases to χE which is integrable (µ(E) < ∞). Thus, by Step 2, χE ∈ F. m+n (e) Suppose E is a Gδ finite measure subset of R . The E = ∞ ∩k=1Uk. Also, there exists an open set U such that µ(U) < ∞ k and E ⊂ U. Set Vk := U ∩ (∩i=1Uk). Then Vk is a decreasing ∞ sequence of open sets such that E = ∩k=1Vk. Thus, χVk is a

decreasing sequence converging to χE. Since χVk ∈ F, by Step 2, χE ∈ F.

(f) Suppose E is of zero measure. Then there exists a Gδ set G such that E ⊂ G and µ(G) = 0. Since χG ∈ F, we have Z Z  Z y χG(x) dx dy = χG(x, y) dx dy = 0. Rn Rm Rm+n

Therefore R χy (x) dx = 0 for a.e. y ∈ n. Thus Gy := {x ∈ Rm G R Rm | (x, y) ∈ G} is of measure zero. Since Ey ⊂ Gy, we have µ(Ey) = 0. Thus, R χy (x) dx = 0 for a.e. y ∈ n. Hence, Rm E R Z Z Z  y χE(x, y) dx dy = χE(x) dx dy = 0. Rm+n Rn Rm CHAPTER 3. LEBESGUE INTEGRATION 78

(g) Suppose E ⊂ Rm+n is a measurable subset with a finite measure. Then there exists a finite measure Gδ set G such that E ⊂ G and µ(G \ E) = 0. Consequently, χE = χG − χG\E, a finite linear combination of functions from F. Thus, χE ∈ F.

Step 4 If f ∈ L1Rm+n then f = f + − f − and an increasing sequence of simple functions converging to f + and f −, respectively. Since simple functions are finite linear combinations. By step 3, χE ∈ F and, by step 1, simple functions belong to F. By step 2, f + and f − are in F and, by step 1 again, f ∈ F.

3.6 Lp Spaces

Recall that we already denoted, in the previous section, the class of integrable functions on Rn as L1(Rn). What was the need for the superscript 1 in the notation?

Definition 3.6.1. For any 0 < p < ∞, a measurable function on E ∈ L(Rn) is said to be p-integrable (Lebesgue) if Z |f(x)|p dx < +∞. E

The space of all Lebesgue p-integrable functions on E ∈ L(Rn) is denoted by Lp(E).

In this sense, our integrable functions are precisely the 1-integrable func- tions. Exercise 69. Show that f ∈ Lp(E) then |f|p ∈ L1(E). The p = ∞ case is a generalisation of the uniform in the space of continuous bounded functions.

Definition 3.6.2. A function f (not necessarily measurable) on E ∈ L(Rn) is said to be essentially bounded if there exists a 0 < M < ∞ such that |f(x)| ≤ M a.e. in E, i.e., the set

n {x ∈ R | |f(x)| > M} CHAPTER 3. LEBESGUE INTEGRATION 79 has outer measure zero. The infimum of all such M is said to be the essential supremum of f. The class of measurable essentially bounded function is denoted by L∞(E).

Exercise 70. Show that Lp(E) forms a vector space over R (or C) for 0 < p ≤ ∞.

Proof. The case p = ∞ is trivial. Consider the case 1 < p < ∞. The closure under scalar multiplication is obvious. For closure under vector addition, we note that |f(x) + g(x)| ≤ |f(x)| + |g(x)| and hence |f(x) + g(x)|p ≤ (|f(x)| + |g(x)|)p for all p > 0. Let 0 < p < ∞ and a, b ≥ 0. Assume wlog that a ≤ b (else we swap their roles). Thus, a + b ≤ 2b = 2 max(a, b) and therefore

(a + b)p ≤ 2pbp ≤ 2p(ap + bp).

Using this we get

(|f(x)| + |g(x)|)p ≤ 2p|f(x)|p + 2p|g(x)|p.

Therefore, Z Z Z |f(x) + g(x)|p ≤ 2p |f(x)|p + 2p |g(x)|p < +∞.

We introduce the notion of “length”, called norm, on Lp(E) for all 0 < p ≤ ∞.

Definition 3.6.3. For all 0 < p < ∞, we define the norm of f ∈ Lp(E) as

Z 1/p p kfkp := |f(x)| dx , E which is finite. For any f ∈ L∞(E), we define its norm as

kfk∞ := inf{M | |f(x)| ≤ M a.e.}. M CHAPTER 3. LEBESGUE INTEGRATION 80

Sometimes it is common to write kfkp as kfkE,p indicating the domain of f, however we shall restrain from complicating our notation and keep track of the domain of f wherever necessary. Let us now observe some properties of the norm. Observe that kfkp ≥ 0. Exercise 71. Show that for each scalar λ ∈ R (or C),

p kλfkp = |λ|kfkp ∀f ∈ L (E).

(This is the reason for having the exponent 1/p in the definition of norm) Note that the norm of zero function, f ≡ 0 is zero, but the converse is not true.

Exercise 72. For each 0 < p ≤ ∞, show that kfkp = 0 iff f = 0 a.e. Observe from the above exercise that the “length” we defined is short of being a “real length” (usually called semi-norm). In other words, we have non-zero vectors whose length is zero. To fix this issue, we inherit the equivalence relation of M(Rn) defined in Definition 2.4.5 to Lp(Rn). Thus, in the quotient space Lp(E)/ ∼ length of all non-zero vectors is non-zero. In practice we always work with the quotient space Lp(E)/ ∼ but write it as Lp(E). Hence the remark following Definition 2.4.5 holds true for Lp(E) (as the quotient space). It now remains to show the triangle inequality of the norm. Proving tri- inequality is a problem due to the presence of the exponent 1/p (which was introduced for dilation property). For instance, the triangle inequality is true without the exponent 1/p in the definition of norm. Exercise 73. Let E ∈ L(Rn). Show that for 0 < p < 1 and f, g ∈ Lp(E) we have p p p kf + gkp ≤ kfkp + kgkp. Proof. Let 0 < p < 1 and a, b ≥ 0. Assume wlog that a ≤ b (else swap their roles). For a fixed p ∈ (0, 1), the function xp satisfies the hypotheses of MVT in [b, a + b] and hence

(a + b)p − bp = pcp−1a for some c ∈ (b, a + b).

Now, since p − 1 < 0, we have

(a + b)p = bp + pcp−1a ≤ bp + pbp−1a ≤ bp + pap−1a = bp + ap. CHAPTER 3. LEBESGUE INTEGRATION 81

Thus, (a + b)p ≤ ap + bp. Using this we have

(|f(x)| + |g(x)|)p ≤ |f(x)|p + |g(x)|p.

Therefore, Z Z Z |f(x) + g(x)|p ≤ |f(x)|p + |g(x)|p < +∞.

Note that we have not proved kf +gkp ≤ kfkp +kgkp, which is the triangle inequality. In fact, triangle inequality is false.

Example 3.10. Let f = χ[0,1/2) and g = χ[1/2,1] on E = [0, 1] and let p = −(1/p) 1/2 < 1. Note that kf + gkp = 1 but kfkp = kgkp = 2 and hence 1−1/p kfkp + kgkp = 2 < 1. Thus, kf + gkp > kfkp + kgkp. Exercise 74. Show that for 0 < p < 1 and f, g ∈ Lp(E),

(1/p)−1 kf + gkp ≤ 2 (kfkp + kgkp) .

This is called the quasi-triangle inequality. What is happening in reality is that the best constant for triangle inequal- ity is max(2(1/p)−1, 1), for all p > 0. Thus, when p > 1 the maximum is 1 and we have the triangle inequality of the norm for p ≥ 1, called the Minkowski inequality. To prove this, we would need the general form of Cauchy-Schwarz inequality, called H¨older’sinequality. For each 1 < p < ∞, we associate with it a conjugate exponent q such that 1/p + 1/q = 1. If p = 1 then we set q = ∞ and vice versa.

Theorem 3.6.4 (H¨older’sInequality). Let E ∈ L(Rn) and 1 ≤ p ≤ ∞. If f ∈ Lp(E) and g ∈ Lq(E), where the q is the conjugate exponent correspond- ing to p, then fg ∈ L1(E) and

kfgk1 ≤ kfkpkgkq. (3.6.1) CHAPTER 3. LEBESGUE INTEGRATION 82

Proof. If either f or g is a zero function a.e then the result is trivially true. Therefore, we assume wlog that both f and g have non-zero norm. Let p = 1 and f ∈ L1(E) and g ∈ L∞(E). Consider Z Z |fg| ≤ ess supx∈E|g(x)| |f| = kgk∞kfk1 < +∞ E E Thus, fg ∈ L1(E). Let 1 < p < ∞ and f ∈ Lp(E) and g ∈ Lq(E). If either kfkp = 0 or kgkq = 0, then equality holds trivially. Thus, we assume wlog that both kfk , kgk > 0. Set f = 1 f ∈ Lp(E) and g = 1 g ∈ Lq(E) p q 1 kfkp 1 kgkq with kf1kp = kg1kq = 1. Recall the AM-GM inequality (cf. (??)), xp yq xy ≤ + . p q Using this we get 1 1 |f (x)g (x)| ≤ |f (x)|p + |g (x)|q 1 1 p 1 q 1

1 1 1 p 1 q |f(x)| |g(x)| ≤ p |f(x)| + q |g(x)| . kfkp kgkq pkfkp qkgkq Now, integrating both sides w.r.t the Lebesgue measure, we get Z   1 p 1 q |fg| ≤ kfkpkgkq p kfkp + q kgkq pkfkp qkgkq

= kfkpkgkq. Hence fg ∈ L1(E). Remark 3.6.5. Equality holds in (3.6.1) iff equality holds in (??) which |f(x)|p |g(x)|q p happens, by Theorem ??, iff p = q , for a.e. x ∈ E. Thus, |f(x)| = kfkp kgkq p q kfkp λ|g(x)| where λ = q . kgkq Exercise 75. Show that for 0 < p < 1, f ∈ Lp(E) and g ∈ Lq(E) where the the conjugate exponent of p (now it is negative),

kfgk1 ≥ kfkpkgkq.

Theorem 3.6.6 (Minkowski Inequality). Let E ∈ L(Rn) and 1 ≤ p ≤ ∞. If f, g ∈ Lp(E) then f + g ∈ Lp(E) and

kf + gkp ≤ kfkp + kgkp. (3.6.2) CHAPTER 3. LEBESGUE INTEGRATION 83

Proof. The proof is obvious for p = 1, since |f(x) + g(x)| ≤ |f(x)| + |g(x)|. Let 1 < p < ∞ and q be the conjugate exponent of p. Observe that Z Z |f(x) + g(x)|p = |f(x) + g(x)|p−1|f(x) + g(x)| Z Z ≤ |f(x) + g(x)|p−1|f(x)| + |f(x) + g(x)|p−1|g(x)|

p−1 p−1 ≤ (f + g) q kfkp + (f + g) q kgkp p−1 ≤ (f + g) q (kfkp + kgkp) Z 1/q p = |f(x) + g(x)| (kfkp + kgkp)

p p/q kf + gkp = kf + gkp (kfkp + kgkp)

kf + gkp = kfkp + kgkp.

Hence f + g ∈ Lp(E).

Exercise 76. Show that for 0 < p < 1 and f, g ∈ Lp(E) such that f, g are non-negative

kf + gkp ≥ kfkp + kgkp The triangle inequality fails for 0 < p < 1 due to the presence of the exponent 1/p in the definition of kfkp. Thus, for 0 < p < 1, we also have the option of ignoring the 1/p exponent while defining kfkp. Define the metric p n p n P n dp : L (R ) × L (R ) → [0, ∞) on L (R ) such that dp(f, g) = kf − gkp for p 1 ≤ p ≤ ∞ and dp(f, g) = kf − gkp for 0 < p < 1. p n Exercise 77. Show that dp is a metric on L (R ) for p > 0.

n Definition 3.6.7. Let E ∈ L(R ). We say a sequence {fk} converges to f p in L (E), p > 0, if dp(fk, f) → 0 as k → ∞.

p µ Exercise 78. Show that if fk → f in L then fk → f in measure. The converse is not true, in general. Moreover, Lp convergence does not imply almost uniform or point-wise a.e. However, they are true for a subsequence. p Exercise 79. If fk converges to f in L (E) then there exists a subsequence

{fkl } of {fk} such that fkl (x) → f(x) point-wise for a.e. x ∈ E. CHAPTER 3. LEBESGUE INTEGRATION 84

Theorem 3.6.8 (Riesz-Fischer). For p > 0, Lp(Rn) is a complete . In general, we ignore studying Lp for 0 < p < 1 because its dual3 is trivial vector space. Thus, henceforth we restrict ourselves to 1 ≤ p ≤ ∞. δ p Exercise 80. Let 1 ≤ p < ∞. For what values of δ ∈ R does |x| ∈ L (B1(0)) n where B1(0) is the unit ball of R . Proof. Consider Z Z Z 1 |x|δp dx = rδp+n−1 dr dσ. B1(0) S1(0) 0

−n Thus, for δp + n − 1 > −1 or δ > p , the integral is finite and is equal to ωn δp+n , where ωn is the surface measure of the unit ball. Also, note that, for −n δ all p < δ < 0, |x| has a blow-up near 0. Proposition 3.6.9. Let E ∈ L(Rn) be such that µ(E) < +∞. If 1 ≤ p < r < ∞ then Lr(E) ⊂ Lp(E) and

µ(E)1/p kfk ≤ kfk . p µ(E)1/r r Proof. Let f ∈ Lr(E). We need to show that f ∈ Lp(E). Set F = |f|p and G = 1. Note that F ∈ Lr/p(E) since Z Z |F |r/p = |f|r < +∞.

Applying H¨older’sinequality, we get

p kfkp = kFGk1 ≤ kF kr/pkGkr/(r−p) p (r−p)/r = kfkrµ(E) 1/p−1/r kfkp ≤ µ(E) kfkr.

Exercise 81. The inclusion obtained√ above is, in general, strict. For instance, observe from Exercise 80 that 1/ x on [0, 1] is in L1([0, 1]) but is not in L2([0, 1]). 3will be introduced in a course of CHAPTER 3. LEBESGUE INTEGRATION 85

Proposition 3.6.10. Let E ∈ L(Rn) be such that µ(E) < +∞. If f ∈ L∞(E) then f ∈ Lp(E) for all 1 ≤ p < ∞ and

kfkp → kfk∞ as p → ∞.

Proof. Given f ∈ L∞(E), note that

Z 1/p p 1/p kfkp = |f| ≤ kfk∞µ(E) . E

1/p Since µ(E) → 1 as p → ∞, we get lim supp→∞ kfkp ≤ kfk∞. Recall that kfk∞ is the infimum over all essential bounds of f. Thus, for every ε > 0 there is a δ > 0 such that µ(Eδ) ≥ δ where

Eδ := {x ∈ E | |f(x)| ≥ kfk∞ − ε}.

Thus, Z p p p kfkp ≥ |f| ≥ (kfk∞ − ε) δ. Eδ 1/p Since δ → 1 as p → ∞, we get lim infp→∞ kfkp ≥ kfk∞ − ε. Since choice of ε is arbitrary, we have limp→∞ kfkp = kfk∞.

∞ n Exercise 82. Let {fk} be a sequence of functions in L (R ). Show that n kfk − fk∞ → 0 iff there is a set E ∈ L(R ) such that µ(E) = 0 and fk → f uniformly on Ec.

Proof. It is enough to show the result for f ≡ 0. Let kfkk∞ → 0. Let n ∞ Ek := {x ∈ R | |fk(x)| > kfkk∞}. Note that µ(Ek) = 0. Set E = ∪k=1Ek. By sub-additivity of Lebesgue measure, µ(E) = 0. Fix ε > 0. Then there c is a K ∈ N such that for all k ≥ K, kfkk∞ < ε. Choose any x ∈ E . Then c |fk(x)| ≤ kfkk∞ for all k. Thus, for all x ∈ E and k ≥ K, |fk(x)| < ε.. c Thus, fk → 0 uniformly on E . n Conversely, let E ∈ L(R ) be such that µ(E) = 0 and fk → 0 uniformly on Ec. Fix ε > 0. For any x ∈ Ec, there exists a K ∈ N (independent of x) such that |fk(x)| < ε for all k ≥ K. For each k ≥ K,

kfkk∞ = ess supx∈Rn |fk(x)| = sup |fk(x)| < ε. x∈Ec

Hence kfkk∞ → 0. CHAPTER 3. LEBESGUE INTEGRATION 86

We now prove some density results of Lp spaces. Recall that in Theo- rem 2.4.11 and Theorem 2.4.12 we proved the density of simple function in M(Rn) under point-wise a.e. convergence. We shall now prove the density of simple functions in Lp spaces. We say a collection of functions A ⊂ Lp is dense in Lp if for every f ∈ Lp and ε > 0 there is a g ∈ A such that kf − gkp < ε.

Theorem 3.6.11. Let E ∈ L(Rn). The class of all simple 4 functions are dense in Lp(E) for 1 ≤ p < ∞.

Proof. Fix 1 ≤ p < ∞ and let f ∈ Lp(E) such that f ≥ 0. By Theorem 2.4.11 we have an increasing sequence of non-negative simple functions {φk} that converge point-wise a.e. to f and φk ≤ f for all k. Thus,

p p p |φk(x) − f(x)| ≤ 2 |f(x)| and by DCT we have Z p p lim kφk − fkp = lim |φk − f| → 0. k→∞ k→∞ E For an arbitrary f ∈ Lp(E), we use the decomposition f = f + − f − where + − f , f ≥ 0. Thus we have sequences of simple functions {φk} and {ψk} such p that φm − ψm → f in L (E) (using triangle inequality). Thus, the space of simple functions is dense in Lp(E).

Theorem 3.6.12. Let E ∈ L(Rn). The space of all compactly supported p continuous functions on E, denoted as Cc(E) is dense in L (E) for 1 ≤ p < ∞.

Proof. It is enough to prove the result for a characteristic function χF , where F ⊂ E such that F is bounded. By outer regularity, for a given ε > 0 there

4By our definition, simple function is non-zero on a finite measure. A simple function φ is a non-zero function on Rn having the (canonical) form

k X φ(x) = ai1Ei i=1

n with disjoint measurable subsets Ei ⊂ R with µ(Ei) < +∞ and ai 6= 0, for all i, and ai 6= aj for i 6= j. CHAPTER 3. LEBESGUE INTEGRATION 87 is an open (bounded) set Ω such that Ω ⊃ F and µ(Ω \ F ) < ε/2. Also, by inner regularity, there is a compact set K ⊂ F such that µ(F \K) < ε/2. By Urysohn lemma there is a continuous function g : E → R such that g ≡ 0 on E \ Ω, g ≡ 1 on K and 0 ≤ g ≤ 1 on Ω \ K. Note that g ∈ Cc(E). Therefore, Z Z p p p kχF − gkp = |χF − g| = |χF − g| ≤ µ(Ω \ K) = ε. E Ω\K

Aliter. Let f ∈ Lp(E) and fix ε > 0. By Theorem 3.6.11, there is a simple function φ such that kφ − fkp < ε/2. Note that φ is supported on a finite measure set, by definition of simple funciton. Let F := supp(φ) and F ⊂ E. By Luzin’s theorem, there is a closed subset Γ ⊂ F such that φ ∈ C(Γ) and

 ε p µ(F \ Γ) < . 2kφk∞

Γ being a closed subset of finite measure set F , Γ is compact in E. Thus, we c put φ to be zero on Γ := E \ Γ, call it g, and g ∈ Cc(E) with supp(g) = Γ. Further, by our construction, we have |g(x)| ≤ kφk∞. Hence, ε ε kg − φkp = kφkp,Γc = kφkp,F \Γ < kφk∞ = . 2kφk∞ 2

p Therefore, kg − fkp < ε. Thus, Cc(E) is dense in L (E).

Example 3.11. The class of all simple functions is not dense in L∞(E). The ∞ space Cc(E) is not dense in L (E), but is dense C0(E) with uniform norm, the space of all continuous function vanishing at infinity.

Theorem 3.6.13. For 1 ≤ p < ∞, Lp(Rn) is separable but L∞(Rn) is not separable.

3.7 Invariance of Lebesgue Integral

Recall that in section 15, we noted the invariance properties of the Lebesgue measure. In this section, we shall note the invariance properties of Lebesgue integral. CHAPTER 3. LEBESGUE INTEGRATION 88

Definition 3.7.1. For any function f on Rn we define its translation by a n vector y ∈ R , denoted τyf, as

τyf(x) = f(x − y).

Similarly, one can define notion similar to reflection

fˇ(x) = f(−x).

Also, dilation by λ > 0, is f(λx).

Exercise 83. Show that if f ∈ L1(Rn)

1 n n (i)( Translation invariance) then τyf ∈ L (R ), for every y ∈ R , and R R f = τyf.

1 n R R (ii)( Reflection) then fˇ ∈ L (R ) and f = fˇ.

1 n R n R (iii)( Dilation) and λ > 0, then f(λx) ∈ L (R ) and f = λ f(λx). Chapter 4

Duality of Differentiation and Integration

The aim of this chapter is to identify the general class functions (within the framework of concepts developed in previous chapters) for which following is true: 1. (Derivative of an integral) d Z x f(t) dt = f(x) dx a

2. (Integral of a derivative)

Z b f 0(x) dx = f(b) − f(a) a

We shall attempt to answer these questions in one-dimensional case to keep our attempt simple. In fact answering both these questions have far- reaching consequences not highlighted in this chapter. Let f :[a, b] → R be Lebegue integrable and define Z x F (x) = f(t) dt x ∈ [a, b]. a Answering first question is equivalent to saying F 0(x) = f(x). Note that for a non-negative f, F is a monotonically increasing function. This observation motivates the study of monotone functions in the next section.

89 CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 90

4.1 Monotone Functions

Recall that a function is said to be monotone if it preserves a given order. A function f is said to be monotonically increasing if f(x) ≤ f(y) whenever x ≤ y. If f(x) < f(y) whenever x ≤ y, we say f is strictly increasing. We prove in this section that a monotone increasing function is differentiable a.e. A major tool for proving this result is the .

Definition 4.1.1. Let E ⊂ Rn. We say a collection of balls V is a Vitali covering of E if for every ε > 0 and for every x ∈ E, there exist a ball B ∈ V such that x ∈ B and µ(B) < ε.

In the definition above we allow the balls to be open or closed but do not allow degenerate balls consisting of single a point or lower dimensional balls. We now prove the Vitali’s covering lemma which claims that one can extract a finite disjoint sub-cover of the Vitali cover such that it “almost” covers E. The proof is constructive.

Lemma 4.1.2 (Vitali Covering Lemma). Let E ⊂ Rn be an arbitrary subset such that µ?(E) < +∞ and let V be a Vitali covering of E. Then, for every k ε > 0, there is a finite disjoint sub-collection {Bi}1 ⊂ V such that

? k  µ E \ ∪i=1Bi < ε.

Proof. Without loss of generality, we assume that each ball in V is closed because ? k  ? k  µ E \ ∪i=1Bi = µ E \ ∪i=1Bi . Let Ω be an open subset of Rn such that µ(Ω) < +∞ and E ⊂ Ω. We now assume without loss of generality that B ⊆ Ω for all B ∈ V. This is possible at the cost of throwing away some elements of the covering V because µ?(E) < +∞ and balls in V as small as possible. First we pick a ball B1 ∈ V. If E ⊂ B1 then we are done. Else, for each k−1 k ≥ 2, we pick a ball Bk such that Bk ∩ (∪i=1 Bi) = ∅ and µ(Bk) > rk/2, where k−1 rk = sup{µ(B) | B ∩ ∪i=1 Bi = ∅}. B∈V

Note that rk is finite for all k, since rk ≤ µ(Ω) < +∞. If the set over which the supremum is taken is an empty collection for some k, i.e., there is no k−1 k−1 B ∈ V such that B ∩ ∪i=1 Bi = ∅, then we already have E ⊂ ∪i=1 Bi and we CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 91 are done. Otherwise, we have a countable disjoint collection of closed balls Bk such that µ(Bk) > rk/2. Also ∪kBk ⊂ Ω and hence, by monotonicity and additivity of µ, we get ∞ ∞ X +∞ > µ(Ω) ≥ µ(∪k=1Bk) = µ(Bk). k=1 Thus, for any given ε > 0, there is K ∈ N such that ∞ X ε µ(B ) < . k 5n k=K+1

? K K We wish to prove that µ (E \ ∪i=1Bi) < ε. Let x ∈ E \ ∪i=1Bi. Such an x exists because otherwise we would not have countable collection and our process would have stopped at Kth stage. For the chosen x we have a ball K Bx ∈ V such that Bx ∩ ∪i=1Bi = ∅. Note that µ(Bx) ≤ rK . We now claim that Bx intersects Bi for some i > K. Suppose, for every i > K, Bx ∩Bi = ∅, then µ(Bx) ≤ ri → 0 as i becomes large. Thus, Bx intersects Bi for some i > K. Let l be the smallest i > K (or first instance) when Bx meets Bl. Hence µ(Bx) ≤ rl < 2µ(Bl). Thus, rx < 2rl. We claim that Bx ⊂ 5Bl. Let y ∈ Bx ∩ Bl. Then, |x − y| ≤ 2rx, where rx is the radius of Bx. Also, |y − xl| ≤ rl where xl is the centre of Bl and rl is its radius. Thus,

|x − xl| ≤ |x − y| + |y − xl| ≤ 2rx + rl < 4rl + rl = 5rl.

Hence x ∈ 5Bl and Bx ⊂ 5Bl. Therefore, K ∞ E \ ∪i=1Bi ⊂ ∪i=K+15Bi and ∞ ? K ∞ n X µ (E \ ∪i=1Bi) ≤ µ(∪i=K+15Bi) ≤ 5 µ(Bi) < ε. i=K+1

Lemma 4.1.3 (Riesz’s Rising Sun Lemma). Let g :[a, b] → R be a continu- ous function. If E := {x ∈ (a, b) | g(x + h) > g(x) for some h > 0} then E is either empty or open. In the latter case (E open) E is a countable union of disjoint intervals (ak, bk) with g(ak) = g(bk) when a 6= ak. For a = ak, g(a) ≤ g(bk). CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 92

Proof. Note that E is empty iff g is non-increasing. Thus, for a generic continuous function g, E is non-empty. Note that E = ∪y∈(a,b)Ey where Ey := {x ∈ (a, y) | g(x) < g(y)}. By the continuity of g, Ey is open and, hence, E is open. Thus, E can be written as a disjoint union of countably many open intervals (ak, bk). Since ak ∈/ E, we have g(ak) ≥ g(bk). Suppose g(ak) > g(bk), then by the continutiy of g, there is a c ∈ (ak, bk) such that

g(a ) + g(b ) g(c) = k k . 2

Among all possible such c choose the one that is closest to bk. Picking such a closest c to bk is possible. If not, then the accumulation point of such c’s g(ak)+g(bk) should be bk which is not possible because g(bk) 6= 2 . Note that c ∈ E and, hence, there is a d > c such that g(d) > g(c). Since bk ∈/ E, we have g(bk) ≥ g(x) for all x ≥ bk. Since g(d) > g(c) > g(bk), we have d < bk. Also, since g(d) > g(c) > g(bk), by continuity of g, there is a c1 ∈ (d, bk) such that g(c1) = g(c). Thus, c1 contradicts the proximity of c with bk. Hence, the hypothesis g(ak) > g(bk) is false and, thus, g(ak) = g(bk).

Theorem 4.1.4. Let f :[a, b] → R be a monotone function. Then f has at most countably many discontinuity points. Conversely, given any countable set E ⊂ R, there exists a monotone function f : R → R whose set of discontinuity is exactly E.

Proof. Without loss of generality let us assume f is increasing. For each x ∈ (a, b), define the left and right limit (h > 0)

f+(x) := lim f(x + h) and f−(x) := lim f(x − h). h→0 h→0

Then, J(x) := f+(x)−f−(x) ≥ 0 is the jump of f at x. Thus, f is continuous at x iff J(x) = 0. For each n ∈ N, define 1 E := {x ∈ (a, b) | J(x) ≥ }. n n

∞ Note that the set of discontinuity points of f is precisely ∪n=1En. Let I : {x1, x2, . . . , xk} be a finite subset of En such that x1 < x2 < . . . < xk. Since f is increasing we have

f(a) ≤ f−(x1) ≤ f+(x1) ≤ ... ≤ f−(xk) ≤ f+(xk) ≤ f(b) CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 93 and k X X k f(b) − f(a) ≥ [f (x ) − f (x )] = J(x) ≥ . + i − i n i=1 x∈I

Thus, the cardinality of En is at most the integer part of n[f(b) − f(a)]. Conversely, let E be a countable set. If E is finite then construct a monotone linear function in the interval between two discontinuity points. Suppose E = {xn} is countable. For each n ∈ N, define an increasing function fn : R → R by ( 1 − n2 if x < xn fn(x) := 1 n2 if x ≥ xn.

Note that fn is discontinuous only at xn. Define, for all x ∈ R,

∞ X f(x) := fn(x). n=1

1 Since |fn(x)| ≤ n2 for all x ∈ R. The is uniformly convergent and, hence, f is well-defined and continuous at every point on which each fn is continuous. Thus, f is continuous on R \ E. We now prove that f is discontinuous at each point of E. Note that, for each n ∈ N, X f = fn + fi. i6=n

P Since i6=n fi is continuous at xn and fn is not continuous at xn, f is discon- tinuous at xn. Further, f is increasing because it is the pointwise limit of a sequence of increasing functions.

Example 4.1. There exists an increasing function f : R → R that is continu- ous at all irrational points and discontinuous at all rational points. To prove the main result of this section, i.e., every increasing function is differentiable a.e., we need the following , called Dini derivatives.

Definition 4.1.5. For any given function f : R → R, we define the four CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 94

Dini derivatives of f at x as follows: f(x + h) − f(x) D+f(x) = lim sup , h→0+ h f(x + h) − f(x) D+f(x) = lim inf , h→0+ h f(x + h) − f(x) f(x) − f(x − h) D−f(x) = lim sup = lim sup , h→0− h h→0+ h f(x + h) − f(x) f(x) − f(x − h) D−f(x) = lim inf = lim inf . h→0− h h→0+ h Note that these numbers always exist and could be infinity. Also, we + − + always have D f(x) ≥ D+f(x) and D f(x) ≥ D−f(x). If D f(x) = D+f(x) 6= ±∞ then we say the right-hand derivative of f exists at x. Simi- − larly, if D f(x) = D−f(x) 6= ±∞ then we say the left-hand derivative of f + − exists at x. We say f is differentiable at x if D f(x) = D+f(x) = D f(x) = 0 + D−f(x) 6= ±∞ and f (x) = D f(x). Observe that for a increasing function the Dini derivatives are all non-negative. Exercise 84. Show that, for a given f : R → R, if g(x) = −f(−x) for all + − x ∈ R then D g(x) = D f(−x) and D−g(x) = D+f(−x). Proof. Consider g(x + h) − g(x) D+g(x) = lim sup h→0+ h −f(−x − h) + f(−x) = lim sup = D−f(−x). h→0+ h Similarly, g(x) − g(x − h) D−g(x) = lim inf h→0+ h −f(−x) + f(−x + h) = lim inf = D+f(−x). h→0+ h

Theorem 4.1.6. If f :[a, b] → R is a increasing function then f is differ- entiable a.e. in [a, b]. Moreover, the derivative f 0 ∈ L1([a, b]) and Z b f 0 ≤ f(b) − f(a). a CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 95

Proof. Owing to Theorem 4.1.4, we assume without loss of generality that f is continuous. To show that f is differentiable a.e. in [a, b], it is enough to show that (a) D+f(x) < ∞ for a.e. in [a, b]

+ (b) and D f(x) ≤ D−f(x) for a.e. in [a, b].

+ Let En := {x ∈ [a, b] | D f(x) > n} for each fixed n ∈ N. Note that En is measurable and is a decreasing sequence of sets. Also,

+ {x ∈ [a, b] | D f(x) = ∞} = ∩nEn.

Now apply Lemma 4.1.3 to gn(x) := f(x) − nx. Therefore, En ⊂ ∪k(ak, bk) and f(bk) − f(ak) ≥ n(bk − ak). Thus, X 1 X 1 µ?(E ) ≤ (b − a ) ≤ [f(b ) − f(a )] ≤ [f(b) − f(a)]. n k k n k k n k k

? ? + Thus, µ (En) → 0 as n → ∞. Thus µ ({D f(x) = ∞}) = 0. Thus, D+f < ∞ almost everywhere. + Now we shall prove that D f(x) ≤ D−f(x) for a.e. in [a, b]. It is enough to show this part because, by applying this result to g(y) = −f(−y), we − would get D f(−y) ≤ D+f(−y) for a.e. −y ∈ [a, b]. Hence,

+ − D f(x) ≤ D−f(x) ≤ D f(x) ≤ D+f(x)

+ and all are equal since D+f(x) ≤ D f(x). Thus, it is sufficient to show that the set + E := {x ∈ [a, b] | D f(x) > D−f(x)} has outer measure zero. In fact, a similar argument will prove the result for every other combination of Dini derivatives. Let p, q ∈ Q such that p > q and define

+ Ep,q := {x ∈ [a, b] | D f(x) > p > q > D−f(x)}.

? Note that E = ∪p,q∈QEp,q. We will show that µ (Ep,q) = 0 which will imply p>q ? ? that µ (E) = 0. To begin we assume a non-empty Ep,q has µ (Ep,q) 6= 0, for a fixed p, q ∈ Q such that p > q, and arrive at a contradiction. We construct a Vitali cover of Ep,q. For any given ε > 0, by outer regularity, there is an CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 96

? open set Ω ⊃ Ep,q such that µ(Ω) < µ (Ep,q) + ε. For each x ∈ Ep.q, since Ω is open, there is an interval [x − h, x] ⊂ Ω such that

f(x) − f(x − h) < qh.

The collection of all such intervals, for each x ∈ Ep,q, forms a Vitali cover of Ep,q. Therefore, by Vitali covering lemma, we have finite disjoint sub- m collection {Ii}1 from the Vitali cover such that

? m µ (Ep,q \ ∪i=1Ii) < ε.

Therefore, we have

m m m X X X ? (f(xi) − f(xi − hi)) < q hi = q µ(Ii) < qµ(Ω) < q(µ (Ep,q) + ε). i=1 i=1 i=1

m Now let A = Ep,q ∩(∪i=1Int(Ii)) and hence Ep,q = A∪(Ep,q \∪Int(Ii)). Thus, ? ? µ (Ep,q) < µ (A) + ε. We shall now construct a Vitali cover for A in terms of Ii. Note that each y ∈ A is contained in Int(Ii) for some i. Choose k > 0 such that [y, y + k] ⊆ Ii and

f(y + k) − f(y) > pk.

` By Vitali covering lemma, there is finite disjoint collection of intervals {Jj}1 each contained in Ii for some i such that

? ` µ (A \ ∪j=1Jj) < ε.

` ` ? ? Set B = A ∩ (∪j=1Jj) and A = B ∪ (A \ ∪j=1Jj). Hence, µ (A) < µ (B) + ε. Therefore, we have

` ` ` X X X (f(yj + kj) − f(yj)) > p kj = p µ(Jj) = pµ(∪jJj) j=1 j=1 j=1 ? ? ? ≥ pµ (B) > p(µ (A) − ε) > p(µ (Ep,q) − 2ε).

Now, for each fixed i, we sum over all j such that Jj ⊂ Ii to get the inequality X (f(yj + kj) − f(yj)) ≤ f(xi) − f(xi − hi),

Jj ⊆Ii CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 97

due to the increasing nature of f and disjointness of Jj. This implies that ? ? ? p(µ (Ep,q) − 2ε) < q(µ (Ep,q) + ε). Since ε is arbitrary, we have pµ (Ep,q) ≤ ? ? qµ (Ep,q) which will contradict q < p unless µ (Ep,q) = 0. Consequently, µ?(E) = 0 and f is differentiable a.e. in [a, b]. Hence f 0(x) is defined a.e. in [a, b]. Set gk(x) = k (f(x + 1/k) − f(x)) 1 0 such that for all x ≥ b, f(x) = f(b) . Note that gk(x) → f (x) a.e. in [a, b]. 0 Thus, f is measurable, due to the measurability of gk which follows from the measurability of f, a consequence of being a increasing function. Also, 0 since f is increasing gk are non-negative and hence f is non-negative. Using Fatou’s lemma, we have Z b Z b  Z b Z b  0 f ≤ lim inf gk = lim inf k f(x + 1/k) − k f(x) a k→∞ a k→∞ a a Z b+1/k Z b ! = lim inf k f(x) − k f(x) k→∞ a+1/k a Z b+1/k Z a+1/k ! = lim inf k f(x) − k f(x) (f constant for x ≥ b) k→∞ b a Z a+1/k ! = lim inf f(b) − k f(x) k→∞ a ≤ f(b) − f(a)(f is increasing).

Note that the above result also holds true for decreasing functions. Also, observe that for any two increasing functions their sum and difference are also differentiable a.e., but the difference is not necessarily increasing or de- creasing. We wish to classify this class of functions which is the difference of two increasing functions.

4.2 Bounded Variation Functions

The problem of finding area under a graph lead to the notion of integration. An equally important problem is to find the length of . Let γ denote 1One could have chosen f(x) = c, for any c ≥ f(b), for all x ≥ b, and obtain c − f(a) but f(b) − f(a) is the best bound one can obtain CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 98 a continuous in a metric space (X, d). Let the continuous function γ :[a, b] → X be the parametrisation of the curve γ with parametrised variable t ∈ [a, b]. Let P be the partition of the interval [a, b], a = t0 ≤ t1 ≤ ... ≤ tk = b. Definition 4.2.1. The length of a curve γ :[a, b] → X on a metric space (X, d) is defined as

( k ) X L(γ) := sup d(γ(ti) − γ(ti−1)) , P i=1 where the supremum is taken over all finite number of partitions P of [a, b]. If L(γ) < +∞ then the curve γ is said to be rectifiable.

The length of the curve is defined as the supremum over the sum of length of all finite number of “line segments” approximating γ. If X is the usual Euclidean space with standard metric then the length of the curve has the form ( k ) X 2 2 1/2 L(γ) = sup |[(γ(ti)) − (γ(ti−1)) ] . P i=1 A interesting questions one can ask at this juncture is: under what conditions on the function γ is the curve γ rectifiable? The length of a curve definition motivates the class of bounded variation functions.

Definition 4.2.2. Let f :[a, b] → R(C) be any real or complex valued func- 2 tion. Let P be a partition of the interval [a, b], a = x0 ≤ x1 ≤ ... ≤ xk = b. We define the of f on [a, b], denoted as V (f;[a, b]), as

( k ) X V (f;[a, b]) := sup |f(xi) − f(xi−1)| . P i=1 We say f is of bounded variation if V (f;[a, b]) < +∞ and the class of all bounded variation function is denoted as BV ([a, b]).

Comparing the definition of bounded variation with curves in C, we ex- pect that any curve γ is rectifiable iff γ :[a, b] → C is a function of bounded variation. 2not necessarily continuous as required for a curve γ CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 99

Example 4.2. Every constant function on [a, b] belongs to BV ([a, b]) and its total variation, V (f;[a, b]) = 0, is zero.

Lemma 4.2.3. For any function f, V (f;[a, b]) = 0 iff f is a constant func- tion on [a, b]

Example 4.3. Any increasing function f on [a, b] has the total variation V (f;[a, b]) = f(b) − f(a). Consequently, if f is a bounded increasing func- tion, then f ∈ BV ([a, b]). For any partition P = {a = x0 ≤ ... ≤ xk = b}, we have

k k X X |f(xi) − f(xi−1)| = (f(xi) − f(xi−1)) i=1 i=1 = f(x1) − f(a) + f(x2) − f(x1) + ... +

+ f(xk−1) − f(xk−2) + f(b) − f(xk−1) = f(b) − f(a).

Thus, V (f;[a, b]) = f(b) − f(a). Similarly, if f is decreasing on [a, b] then f ∈ BV ([a, b]) and V (f;[a, b]) = f(a) − f(b). The above example is very important and we will later see that every function of bounded variation can be decomposed in to increasing bounded functions, a result due to Jordan.

Example 4.4. The Cantor function fC on [0, 1] is increasing and hence is in BV ([0, 1]). We already know fC is uniformly continuous (cf. AppendixA). Example 4.5. Any differentiable function f :[a, b] → R such that f 0 is bounded (say by C) is in BV ([a, b]). Using , we know that |f(x) − f(y)| f 0(z) = ∀x, y ∈ [a, b] and z ∈ [x, y]. |x − y| Since the derivative is bounded, we get |f(x) − f(y)| ≤ C|x − y|. Thus, for any partition P = {a = x0 ≤ ... ≤ xk = b}, we have

k k X X |f(xi) − f(xi−1)| ≤ C |xi − xi−1| = C(b − a). i=1 i=1 The above example is a particular case of the class of Lipschitz functions on [a, b]. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 100

Definition 4.2.4. A function f :[a, b] → R(C) is said to be Lipschitz on [a, b] if there exists a Lipschitz constant C > 0 such that

|f(x) − f(y)| ≤ C|x − y| ∀x, y ∈ [a, b].

The space of all Lipschitz functions is denoted as Lip([a, b]).

Exercise 85. Any Lipschitz function is uniformly continuous, Lip([a, b]) ⊂ C([a, b]) Exercise 86. Every Lipschitz function is of bounded variation and

V (f;[a, b]) ≤ C(b − a), i.e., Lip([a, b]) ⊂ BV ([a, b]). Exercise 87. Every element of BV ([a, b]) is a bounded function on [a, b].

Exercise 88. Show that the characteristic function f = χQ∩[a,b] on [a, b] do not belong to BV ([a, b]).

Proof. The idea behind the proof is that for each fixed n ∈ N, we shall construct a partition P = {x0, x1, . . . , xn, xn+1, xn+2} of [a, b] such that

n+2 X V (f;[a, b]) ≥ |f(xi) − f(xi−1)| > n. i=1

Choose x0 = a. Let x1 be an irrational between a and b. Choose x2 to be an rational between x1 and b. Proceeding this way till xn+2 = b, we will have a partition P whose successive points, excluding a and b, alternate between rational and irrational. Therefore,

n+2 X V (f;[a, b]) ≥ |f(xi) − f(xi−1)| i=1 n+1 X ≥ |f(xi) − f(xi−1)| i=2 = |f(x2) − f(x1)| + ... + |f(xn+1) − f(xn)| = |1 − 0| + |0 − 1| + ... = n.

Thus, V (f;[a, b]) = ∞. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 101

We have already seen that the Cantor function fC , which is (uniformly) continuous, is of bounded variation. But we do have functions which are continuous and not of bounded variation. Exercise 89. Show that the following continuous function on [0, 1] ( x sin(1/x) x 6= 0 f(x) = 0 x = 0 do not belong to BV ([0, 1]). More generally, ( xα sin(1/xβ) x 6= 0 g(x) = 0 x = 0 is in BV ([0, 1]) iff α > β.

Proof. For each k ∈ N, note that  1   1  (−1)k f = 0 and f = . kπ kπ + π/2 kπ + π/2

1 If we choose the points in our partition alternating between 1/kπ and kπ+π/2 then, for each m ∈ N,

m k X (−1) V (f;[a, b]) ≥ 0 − . kπ + π/2 k=1 Hence, ∞ X 1 V (f;[a, b]) ≥ = ∞. kπ + π/2 k=1 For each k ∈ N, note that  1   1  (−1)k g = 0 and f = . (kπ)1/β (kπ + π/2)1/β (kπ + π/2)α/β

If we choose the points in our partition alternating between 1/(kπ)1/β and 1 (kπ+π/2)1/β then, for each m ∈ N,

m k X (−1) V (f;[a, b]) ≥ 0 − . (kπ + π/2)α/β k=1 CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 102

Hence, ∞ X 1 V (f;[a, b]) ≥ . (kπ + π/2)α/β k=1 The series on the right converges iff α/β > 1. Thus, g ∈ BV ([0, 1]) implies α > β. The converse part needs a proof.

Lemma 4.2.5. Let f :[a, b] → R be a given function. Let P denote the parti- 0 tion P = {a, x1, . . . , xn−1, b} of the interval [a, b] and P = {a, y1, . . . , ym−1, b} be a refinement of P , i.e., P ⊂ P 0. Then X X |f(xi) − f(xi−1)| ≤ |f(yj) − f(yj−1)|. P P 0 Proof. We first prove the result by adding one point to the partition P and 0 then invoke induction. Let y ∈ P . If y = xi, for some i, then the partition P remains unchanged. If y 6= xi for all i then y ∈ (xk−1, xk) for some k ∈ {0, 1, . . . , n}. Consider,

k−1 X X |f(xi) − f(xi−1)| = |f(xi) − f(xi−1)| + |f(xk) − f(xk−1)| P i=1 n X + |f(xi) − f(xi−1)| i=k+1 k−1 n X X = |f(xi) − f(xi−1)| + |f(xi) − f(xi−1)| i=1 i=k+1

+ |f(xk) − f(y) + f(y) − f(xk−1)| k−1 X ≤ |f(xi) − f(xi−1)| + |f(xk) − f(y)| i=1 n X + |f(y) − f(xk−1)| + |f(xi) − f(xi−1)| i=k+1 n+1 X = |f(xi) − f(xi−1)| (by relabelling). i=1 Similarly, adding each point of P 0 into the extended partition of P , we have our result. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 103

Exercise 90. Show that BV ([a, b]) forms a vector space over R. Also, if f, g ∈ BV ([a, b]) then (i) fg are in BV ([a, b]).

(ii) f/g ∈ BV ([a, b]) if 1/g is bounded on [a, b].

Theorem 4.2.6. Let f :[a, b] → R be a function and let c ∈ (a, b). If f belongs to both BV ([a, c]) and BV ([c, b]) then f ∈ BV ([a, b]) and

V (f;[a, b]) = V (f;[a, c]) + V (f;[c, b]).

Proof. Let P be any partition of [a, b] and P 0 = P ∪ {c}, relabelled in in- creasing order. P 0 being a refinement of P , arguing similar to the proof of Lemma 4.2.5, we get X X |f(xi) − f(xi−1)| ≤ |f(xi) − f(xi−1)| P P 0 X X = |f(xi) − f(xi−1)| + |f(xi) − f(xi−1)|

xi≤c xi≥c ≤ V (f;[a, c]) + V (f;[c, b]).

Hence, V (f;[a, b]) ≤ V (f;[a, c]) + V (f;[c, b]). On the other hand, let P1 and P2 be a partition of [a, c] and [c, b], respectively. Then P = P1 ∪ P2 gives a partition of [a, b]. Therefore, X X X |f(xi) − f(xi−1)| + |f(xi) − f(xi−1)| = |f(xi) − f(xi−1)|

P1 P2 P ≤ V (f;[a, b]).

The above inequality is true for any arbitrary partition P1 and P2 of [a, c] and [c, b], respectively, Thus,

V (f;[a, c]) + V (f;[c, b]) ≤ V (f;[a, b]) and we have equality as desired. Exercise 91. Show that if f ∈ BV ([a, b]) then f ∈ BV ([c, d]) for all sub- intervals [c, d] ⊂ [a, b].

Let f : R → R(C) be any real or complex valued function. Let BVloc(R) denote the class of all f : R → R such that f ∈ BV ([a, b]) for all [a, b] ⊂ R. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 104

Definition 4.2.7. We define the total variation of f on R, denoted as V (f; R), as V (f; R) := sup V (f;[a, b]), [a,b] where the supremum is taken over all closed intervals [a, b] ⊂ R. We say f is of bounded variation on R if V (f; R) < +∞ and denote the class as BV (R).

Note that BV (R) ⊂ BVloc(R) and the inclusion is strict. Example 4.6. The function f ( 1 x 6= 1 f(x) = 1−x 0 x = 1 belongs to BV (0, 1) but do not belong to BV ([0, 1]). Definition 4.2.8. For f ∈ BV ([a, b]), we define its variation function, ( V (f;[a, x]) ∀x ∈ (a, b] Vf (x) = 0 x = a

Lemma 4.2.9. The variation function Vf (x) corresponding to a function f ∈ BV ([a, b]) is an increasing function.

Proof. Let x, y ∈ [a, b] be such that x < y. We claim that Vf (x) ≤ Vf (y). By, Theorem 4.2.6, we have

V (f;[a, y]) = V (f;[a, x]) + V (f;[x, y]) V (f;[a, y]) − V (f;[a, x]) = V (f;[x, y])

Vf (y) − Vf (x) = V (f;[x, y]).

Since V (f;[x, y]) ≥ 0, we have Vf (y) ≥ Vf (x) and equality holds when f is constant on [x, y].

Theorem 4.2.10 (Jordan Decomposition). Let f :[a, b] → R be a real valued function. Then the following are equivalent: (i) f ∈ BV ([a, b])

(ii) There exist two increasing functions f1, f2 :[a, b] → R such that f = f1 − f2. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 105

Proof. (ii) implying (i) is obvious, because any increasing function is in the vector space BV ([a, b]). Conversely, let us prove (i) implies (ii). For a given f ∈ BV ([a, b]) we know that Vf , the variation function, is increasing in [a, b]. Set f1 = Vf and f2 = Vf − f. It only remains to show that f2 is increasing. Let x, y ∈ [a, b] be such that x < y. Consider,

f2(y) − f2(x) = Vf (y) − f(y) − Vf (x) + f(x)

= Vf (y) − Vf (x) − (f(y) − f(x)) = V (f;[x, y]) − (f(y) − f(x)) ≥ V (f;[x, y]) − |f(y) − f(x)| ≥ 0.

Thus, f2 is increasing and f = f1 − f2. Exercise 92. Show that in the above theorem one can, in fact, have strictly increasing functions f1, f2.

Proof. f = g1 − g2 where g1 := f1 + x and g2 := f2 + x. Theorem 4.2.11 (Lebesgue Differentiation Theorem). If f ∈ BV ([a, b]) then f is differentiable a.e. in [a, b] and the derivative f 0 ∈ L1([a, b]). Fur- ther, Z b |f 0| ≤ V (f;[a, b]). a Proof. The fact that f is differentiable a.e. and f 0 ∈ L1[a, b] follows from the Jordan decomposition (Theorem 4.2.10) and Theorem 4.1.6. Also, by Lemma 4.2.9, Vf is an increasing function. Thus, again by Theorem 4.1.6, Vf is differentiable a.e. and Z b 0 Vf (x) dx ≤ Vf (b) − Vf (a) = Vf (b) = V (f;[a, b]). a For any x, y ∈ [a, b], we have

Vf (y) − Vf (x) = V (f;[x, y]) ≥ |f(y) − f(x)| ≥ f(y) − f(x).

0 0 0 0 0 Thus, f ≤ Vf and |f | ≤ |Vf | = Vf . Therefore, Z b Z b 0 0 |f (x)| dx ≤ Vf (x) dx ≤ V (f;[a, b]). a a CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 106

Exercise 93. Give an example of a function f : [0, 1] → R such that f∈ / BV ([0, 1]) but f is differentiable everywhere in [0, 1]. Proof. Let ( x2 sin(1/x2) x ∈ (0, 1] f(x) = 0 x = 0 then ( 2x sin(1/x2) − 2 cos(1/x2) x ∈ (0, 1] f 0(x) = x 0 x = 0.

The derivative f 0 considered in above exercise is not in L1([0, 1]). The L1 belonging is very crucial to prove the converse of Lebesgue differentiation theorem. A kind of converse of Lebesgue differentiation theorem is proved in Theorem 4.4.5.

4.3 Derivative of an Integral

We now have enough tools to answer the first question posed in the beginning of this chapter. Let f ∈ L1([a, b]) and set Z x F (x) = f(t) dt. a We have already seen in Exercise 68 that F is continuous. We now show that F has bounded variation. In fact, a stronger statement is true, F is absolutely continuous, which we will prove once we introduce the definition of absolute continuity of a function. Lemma 4.3.1. If f ∈ L1([a, b]) then the continuous function F ∈ BV ([a, b]). Proof. Consider

n n n Z xi Z xi Z b X X X |F (xi) − F (xi−1)| = f(t) dt ≤ |f(t)| dt = |f| < ∞. i=1 i=1 xi−1 i=1 xi−1 a Hence V (f;[a.b]) < ∞. Lemma 4.3.2. If f ∈ L1([a, b]) and F ≡ 0 on [a, b] then f = 0 a.e. on [a, b]. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 107

Proof. We shall prove by contradiction. Let E := {x ∈ [a, b] | f(x) > 0}. Assume E is of non-zero measure. By inner regularity there exists a closed set Γ ⊂ E such that µ(E \ Γ) < ε for any given ε > 0. Hence µ(Γ) > 0. Set Ω = (a, b) \ Γ. Since F is identically zero, we have

Z b Z Z 0 = F (b) = f(t) dt = f + f. a Γ Ω Thus, Z Z f = − f 6= 0. Ω Γ ∞ Since Ω is open, Ω = ∪i=1(ai, bi) is a disjoint union of open intervals. Then,

∞ Z X Z bi 0 6= f = f. Ω i=1 ai

Therefore, for some k ∈ N, we have

Z bk Z bk Z ak 0 6= f = f − f = F (bk) − F (ak). ak a a

Thus, either F (bk) 6= 0 or F (ak) 6= 0 which contradicts the hypothesis on F . Similar argument is valid for the set E on which f < 0. Hence proved.

Theorem 4.3.3. Let f ∈ L1([a, b]) and c ∈ R. Set Z x F (x) = c + f(t) dt. a

Then c = F (a) and F 0 = f a.e. on [a, b].

Proof. The fact that c = F (a) is obvious. By Lemma 4.3.1, we have F ∈ BV ([a, b]). By Lebesgue differentiation theorem, F is differentiable a.e. It is required to show that F 0 = f a.e. We prove by cases. First let us assume f is bounded, i.e., kfk∞ < ∞. Extend F as F (x) = F (b), for all x ≥ b. Set

Z x+1/k gk(x) := k (F (x + 1/k) − F (x)) = k f(t) dt. x CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 108

0 Note that gk(x) → F (x) a.e. in [a, b]. Since f is bounded, gk’s are all uniformly bounded and supported inside [a, b]. Using BCT, for any d ∈ [a, b], we have Z d Z d  Z d Z d  0 F = lim gk = lim k F (x + 1/k) − k F (x) a k→∞ a k→∞ a a Z d+1/k Z d ! = lim k F (x) − k F (x) k→∞ a+1/k a Z d+1/k Z a+1/k ! = lim k F (x) − k F (x) . k→∞ d a

Consider, for any e ∈ [a, b],

Z e+h Z e+h 1 1 F (e) − lim F (x) dx ≤ lim |F (e) − F (x)| dx h→0 h e h→0 h e ≤ sup |F (e) − F (x)|. x∈[e,e+h]

By continuity of F (cf. Exercise 68), F (e + h) → F (e) as h → 0. Hence, we have 1 Z e+h lim F (x) dx = F (e). h→0 h e Therefore, Z d Z d F 0 = F (d) − F (a) = f(t) dt. a a Thus, for all d ∈ [a, b], we have

Z d (F 0 − f) = 0 a and by Lemma 4.3.2, we have F 0 − f = 0 a.e. on [a, b], i.e., F 0 = f a.e. on [a, b]. It now remains to prove the result for a unbounded function. Without loss of generality, we assume f is non-negative. Then F is increasing on [a, b] and, by Theorem 4.1.6, we have

Z b F 0(x) dx ≤ F (b) − F (a). a CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 109

Let fk be the truncation of f at k level, i.e., ( f(x) f(x) ≤ k fk(x) = k f(x) > k.

Each fk is a bounded by k and they converge a.e. to f. Also, f − fk is a non-negative function. Define Z x Gk(x) = (f(t) − fk(t)) dt. a

Note that Gk is increasing function on [a, b] and hence is in BV ([a, b]). Thus, 0 Gk is differentiable a.e. Since {fk} are each bounded, we have fk(x) = Fk(x) a.e. where Z x Fk(x) = c + fk(t) dt. a Therefore, 0 0 0 0 Gk(x) = F (x) − Fk(x) = F (x) − fk(x).

Since Gk’s are increasing its derivative is non-negative and hence, we have 0 0 F (x) ≥ fk(x) a.e. Consequently, F (x) ≥ f(x) a.e. Thus,

Z b Z b F 0(x) dx ≥ f(x) dx = F (b) − F (a) a a and we have equality above, since other inequality holds as noted above. Therefore, Z b (F 0(x) − f(x)) dx = 0 a and for F 0 − f ≥ 0, integral zero implies that F 0(x) = f(x) a.e. on [a, b]. Recall the definition of derivative of F ,

F (x + h) − F (x) 1 Z x+h F 0(x) = lim = lim f(t) dt. h→0 h h→0 h x Thus, we may reformulate the demand F 0(x) = f(x) a.e. on [a, b] as

1 Z x+h lim f(t) dt = f(x) for a.e. x ∈ [a, b]. h→0 h x CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 110

Note that the integral (along with the fraction) on the LHS is the “average” or “mean” of f over [x, x+h] and the equality says that the limit of averages of f around a interval I of x converges to the value of f at x, as the measure of interval I tends to zero. The theorem proved above validates this result for all f ∈ L1 in the one dimension case. This reformulation, in terms of averages, helps in stating the problem in higher dimensions. Thus, in higher dimension, we ask the question: For all f ∈ L1(Rn), do we have Z 1 n lim f(t) dt = f(x) for a.e. x ∈ R (4.3.1) µ(B)→0 µ(B) B x∈B where B is any ball in Rn containing x? Note that if f is continuous at x then (4.3.1) holds true because for every ε > 0 there is a δ > 0 such that |f(x) − f(t)| < ε whenever |x − t| < δ and, hence, for any ball B with radius less than δ/2 that contains x, we have Z Z 1 1 f(x) − f(t) dt ≤ |f(x) − f(t)| dt < ε. µ(B) B µ(B) B Theorem 4.3.4 (Lebesgue-Besicovitch Differentiation Theorem). Let µ be n 1 n a on R and f ∈ Lloc(R , µ). Then 1 Z lim f dµ = f(x) h→0 µ(Bh(x)) Bh(x) for µ a.e. x ∈ Rn. For the case when µ is a Lebesgue measure, we define the precise repre- sentative of f as ( lim 1 R f dx if the limit exists f ?(x) := h→0 µ(B) B 0 otherwise

By Lebesgue-Besicovitch differentiation theorem, f = f ? a.e., i.e., they are in the same equivalence class. If f = g a.e then 1 Z 1 Z lim f(t) dt = lim g(t) dt h→0 µ(B) B h→0 µ(B) B whenever the limit exists. Thus, f ? = g? for all x ∈ Rn. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 111

Definition 4.3.5. Let 1 ≤ p < ∞, µ be a Radon measure on Rn and f ∈ p n n Lloc(R , µ). A point x ∈ R is said to be a Lebesgue point of f w.r.t µ if 1 Z lim |f − f(x)|p dµ = 0. h→0 µ(Bh(x)) Bh(x) The set of all Lebesgue points of f w.r.t µ is called the Lebesgue set of f w.r.t µ.

In this terminology, Lebesgue differentiation theorem says that the com- plement of Lebesgue set is of measure zero. Further, the set of all continuity points of f is contained in the Lebesgue set.

Corollary 4.3.6. Let 1 ≤ p < ∞, µ be a Radon measure on Rn and f ∈ p n Lloc(R , µ). The complement of the Lebesgue set of f w.r.t µ has measure zero w.r.t µ.

Corollary 4.3.7. If µ is the Lebesgue measure then result with balls having centre at x is also true for any ball containing x, i.e., Z 1 p n lim |f(t) − f(x)| dt = 0 for a.e. x ∈ R . µ(B)→0 µ(B) B x∈B

4.4 Absolute Continuity and FTC

In this section we wish to identify the class of functions f for which

Z b f 0(x) dx = f(b) − f(a). a This is the second question we hoped to answer in the beginning of the chap- ter. Note that if f ∈ BV ([a, b]) then, by Lebesgue differentiation theorem, f is differentiable and the derivative is Lebesgue integral. So the question reducing to asking: For any f ∈ BV ([a, b]), do we have

Z b f 0(x) dx = f(b) − f(a)? a Unfortunately, the answer is a “No”, as seen in the example below. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 112

Example 4.7. Recall that the Cantor function fC ∈ BV ([0, 1]), since fC is 0 increasing. Outside of the Cantor set C, fC is constant and hence fC = 0 a.e. on [0, 1]. Therefore, Z 1 0 fC = 0 0 but fC (1) − fC (0) = 1 − 0 = 1. This motivates us to look for a sub-class of bounded variation functions for which fundamental theorem of calculus (FTC) is true.

Definition 4.4.1. A function f :[a, b] → R is said to be absolutely contin- uous on [a, b] if for every ε > 0 there exist a δ > 0 such that X |f(yi) − f(xi)| < ε. i for any disjoint collection (finite or countable) of subintervals {(xi, yi)} of [a, b] with X |yi − xi| < δ. i Let AC([a, b]) denote the set of all absolutely continuous functions on [a, b].

Exercise 94. Show that AC([a, b]) forms a vector space over R or C. Also, show that f, g ∈ AC([a, b]) then fg ∈ AC([a, b]). Exercise 95. If f ∈ AC([a, b]) then |f| ∈ AC([a, b]). Exercise 96. Show that AC([a, b]) ⊂ C([a, b]). The inclusion is proper. Show that the Cantor function, which is continuous, fC ∈/ AC([a.b]). Example 4.8. Every constant function on [a, b] is in AC([a, b]). 1 R x Example 4.9. For any f ∈ L ([a, b]), F (x) = a f(t) dt is absolutely contin- uous in [a, b]. Let ε > 0 be given. By Proposition 3.4.6, we have δ > 0 such that Z |f| < ε whenever µ(E) < δ. E

Let us pick a collection of disjoint subintervals {(xi, yi)} ⊂ [a, b] such that P i |yi − xi| < δ. Consider X X Z yi Z |F (yi) − F (xi)| ≤ |f(t)| dt = |f(t)| dt < ε. i i xi ∪i(xi,yi) CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 113

Example 4.10. Lip([a, b]) ⊂ AC([a, b]). This inclusion is proper.

Lemma 4.4.2. Every absolutely continuous function is of bounded varia- tion, i.e., AC([a, b]) ⊂ BV ([a, b]). Consequently, if f ∈ AC([a, b]) then f is differentiable a.e. in [a, b] and f ∈ L1([a, b]).

Proof. Let f ∈ AC([a, b]) and δ > 0 be such that for any disjoint collection (finite or countable) of subintervals {(xi, yi)} of [a, b] with X |yi − xi| < δ i we have X |f(yi) − f(xi)| < 1. i Let M denote the smallest integer such that (b − a)/δ ≤ M. Let P be any partition of [a, b]. We refine the partition P into P 0 such that P 0 has precisely M intervals and hence each of the interval has length less than δ. Therefore, 0 for each subinterval of the partition P , we have |f(xi) − f(xi−1)| < 1. Thus, X |f(xi) − f(xi−1)| < M. P 0 But by Lemma 4.2.5, we have X X |f(xi) − f(xi−1)| ≤ |f(xi) − f(xi−1)| < M. P P 0

Hence, V (f;[a, b]) < M and f ∈ BV ([a, b]).

The inclusion proved above is proper. The Cantor function fC on [0, 1] is in BV ([0, 1] and fC ∈/ AC([0, 1]). We now define a class of functions which are complementary in nature to absolutely continuous functions.

Definition 4.4.3. A function f :[a, b] → R is said to be singular if f is differentiable a.e. in [a, b] and f 0 = 0 a.e. in [a, b].

Example 4.11. The Cantor function fC is singular on [0, 1]. In fact, any non-constant function is either singular or absolutely contin- uous. CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 114

Theorem 4.4.4. If f :[a, b] → R is both absolutely continuous and singular, then f is constant. Proof. It is enough to show that f(a) = f(c) for all c ∈ (a, b]. Fix a c ∈ (a, b]. Due to the singular nature of f, we have a measurable set E ⊂ (a, c) such that µ(E) = c − a and f 0(x) = 0 on E. Due to the absolute continuity of f, for every ε > 0 there exist a δ > 0 such that X ε |f(y ) − f(x )| < . i i 2 i for any disjoint collection (finite or countable) of subintervals {(xi, yi)} of [a, b] with X |yi − xi| < δ. i We now construct a Vitali cover for E. For each x ∈ E, f is differentiable and derivative is zero. We choose the interval [x, x + h] ⊂ [a, c] such that ε |f(x + h) − f(x)| < h. 2(c − a) By Vitali covering lemma, we can find a finite disjoint collection of closed k intervals {Ii = [xi, xi + hi]}1 such that ? k µ (E \ ∪i=1Ii) < δ.

Note that [a, c] is an interval and {xi, xi + hi}. for all i = 1, 2, . . . , k is a disjoint collection of intervals in [a, c]. Thus, relabelling as {x0 = a, x1, x1 + h1, x2, x2 + h2, . . . , xk, xk + hk, c = xk+1} and setting h0 = 0, we have that

k X |xi+1 − (xi + hi)| < δ. i=0 Hence, by absolute continuity, we have

k X ε |f(x ) − f(x + h )| < i+1 i i 2 i=0 Also,

k k X ε X ε ε |f(x + h ) − f(x )| < h < (c − a) = . i i i 2(c − a) i 2(c − a) 2 i=1 i=1 CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 115

Now, consider

k k X X |f(c) − f(a)| = f(xi+1) − f(xi + hi) + f(xi + hi) − f(xi) i=0 i=1 k k X X ≤ |f(xi+1) − f(xi + hi)| + |f(xi + hi) − f(xi)| i=0 i=1 ε ε < + = ε. 2 2 Since the choice of ε is arbitrary, we have |f(c) − f(a)| = 0 and f(c) = f(a).

Theorem 4.4.5. If f :[a, b] → R then the following are equivalent:

(i) f ∈ AC([a, b])

(ii) f is differentiable a.e.,f 0 ∈ L1([a, b]) and

Z x f(x) − f(a) = f 0(t) dt x ∈ [a, b]. a

Proof. We first prove (ii) implies (i). Let ε > 0 be given. By Proposi- tion 3.4.6, since f 0 ∈ L1([a, b]), there exists a δ > 0 such that Z |f 0| < ε whenever µ(E) < δ. E

Consider any disjoint collection of subintervals of [xi, yi] ⊂ [a, b] such that P P i |yi − xi| < δ. We claim that i |f(yi) − f(xi)| < ε. Consider

Z yi Z yi X X 0 X 0 |f(yi) − f(xi)| = f (t) dt ≤ |f | < ε.

i i xi i xi

Conversely, let f ∈ AC([a, b]) then f ∈ BV ([a, b]). Thus, by Lebesgue differentiation theorem, f is differentiable a.e. and f 0 ∈ L1([a, b]). Define

Z x F (x) := f 0(t) dt a CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 116

By Example 4.9, F ∈ AC([a, b]) and hence g = f − F ∈ AC([a, b]). By Theorem 4.3.3, we have F 0 = f 0 a.e. on [a, b]. Thus, g0 = 0 a.e., hence g is singular. Therefore, g is constant, g ≡ c and Z x f(x) = c + F (x) = c + f 0(t) dt. a Thus, c = f(a) and Z x f(x) − f(a) = f 0(t) dt. a

The implication (ii) implies (i) is a kind of converse to Lebesgue differ- entiation theorem (with additional hypothesis). In fact, the exact converse of Lebesgue differentiation theorem, i.e., f 0 exists and f 0 ∈ L1 implies that f ∈ AC ⊂ BV , is true, but requires more observation, viz. Banach-Zaretsky theorem.

Corollary 4.4.6. If f ∈ BV ([a, b]) then f can be decomposed as f = fa + fs where fa ∈ AC([a, b]) and fs is singular on [a, b]). Moreover, fa and fs are unique up to additive constants and Z x 0 f (t) dt = fa(x). a Proof. Since f ∈ BV ([a, b]), by Lebesgue differentiation theorem, f 0 exists and f 0 ∈ L1([a, b]). Define Z b 0 fa := f (x) dx. a 0 0 By (b), fa ∈ AC([a, b]). By, the derivative of an integral we have, fa = f 0 a.e. on [a, b]. Define fs := f − fa. Thus, fs = 0 a.e. and hence fs is singular. Let f = g+h where g ∈ AC([a, b]) and h is singular. Then, g+h = fa+fs. Hence g − fa = fs − h. LHS is in AC([a, b]) and RHS is singular. Therefore they must be equal to some constant c. g = fa + c and h = fs − c.

Exercise 97 (). Show that if f, g ∈ AC([a, b]) then Z b Z b f(x)g0(x) dx + f 0(x)g(x) dx = f(b)g(b) − f(a)g(a). a a CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 117

Theorem 4.4.7. If f ∈ AC([a, b]) then

Z b V (f;[a, b]) = |f 0|. a

Proof. By Lemma 4.4.2, f ∈ BV [a, b]. Thus, by Theorem 4.2.11, R |f 0| ≤ V (f;[a, b]). Consider any partition P := {a, x1, . . . , xn, b} of [a, b]. Since f ∈ AC[xi−1, xi] for all i = 1, . . . , n + 1, we have from Theorem 4.4.5 that

Z xi 0 f(xi) − f(xi−1) = f . xi−1

Thus, Z xi Z b X X 0 0 |f(xi) − f(xi−1)| ≤ |f | = |f | P P xi−1 a

R b 0 and taking supremum over all partitions we establish V (f;[a, b]) ≤ a |f |. Thus, equality holds.

Theorem 4.4.8. Let f ∈ AC([a, b]). Then µ(f(E)) = 0 for all E ⊆ [a, b] such that µ(E) = 0.

Proof. Let E ⊆ (a, b) be such that µ(E) = 0. Note that we are excluding the end-points because {f(a), f(b)} is measure zero subset of f(E). Since f ∈ AC([a, b]), for every given ε > 0, there exists a δ > 0 such that for every P sub-collection of disjoint intervals {(xi, yi)} ⊂ [a, b] with (yi − xi) < δ, P i we have i |f(yi) − f(xi)| < ε. By outer regularity of E, there is an open set Ω ⊃ E such that µ(Ω) < δ. Wlog, we may assume Ω ⊂ (a, b), because otherwise we consider the intersection of Ω with (a, b). But Ω = ∪i(xi, yi), a disjoint countable union of open intervals and X |yi − xi| = µ(Ω) < δ. i

Consider,

? ? ? µ (f(E)) ≤ µ (f(Ω)) = µ (f(∪i(xi, yi))) ? X ? = µ (∪if(xi, yi)) ≤ µ (f(xi, yi)). i CHAPTER 4. DUALITY OF DIFFERENTIATION AND INTEGRATION 118

Let ci, di ∈ (xi, yi) be points such that f(ci) and f(di) is the minimum and maximum, respectively, of f on (xi, yi). Note that |di − ci| ≤ |yi − xi| and P hence i |di − ci| < δ. Then,

X ? X µ (f((xi, yi))) = |f(di) − f(ci)| < ε. i i Thus, µ?(f(E)) = 0 and f(E) is measurable set. Appendices

119

Appendix A

Cantor Set and Cantor Function

Let us construct the Cantor set which plays a special role in analysis. Consider C0 = [0, 1] and trisect C0 and remove the middle open interval to get C1. Thus, C1 = [0, 1/3] ∪ [2/3, 1]. Repeat the procedure for each interval in C1, we get

C2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1].

Repeating this procedure at each stage, we get a sequence of subsets Ci ⊆ [0, 1], for i = 0, 1, 2,.... Note that each Ck is a compact subset, since it is a finite union of compact sets. Moreover,

C0 ⊃ C1 ⊃ C2 ⊃ ... ⊃ Ci ⊃ Ci+1 ⊃ ....

∞ The Cantor set C is the intersection of all the nested Ci’s, C = ∩i=0Ci. Lemma A.0.1. C is compact. Proof. C is countable intersection of closed sets and hence is closed. C ⊂ [0, 1] and hence is bounded. Thus, C is compact. The Cantor set C is non-empty, because the end-points of the closed intervals in Ci, for each i = 0, 1, 2,..., belong to C. In fact, the Cantor set cannot contain any interval of positive length. Lemma A.0.2. For any x, y ∈ C, there is a z∈ / C such that x < z < y. (Disconnected)

121 APPENDIX A. CANTOR SET AND CANTOR FUNCTION 122

Proof. If x, y ∈ C are such that z ∈ C for all z ∈ (x, y), then we have the open interval (x, y) ⊂ C. It is always possible to find i, j such that

 j j + 1 , ⊆ (x, y) 3i 3i but does not belong Ci ⊃ C.

We show in example 2.5, that C has length zero. Since C is non-empty, how ‘big’ is C? The number of end-points sitting in C are countable. But C has points other than the end-points of the closed intervals Ci for all i. For instance, 1/4 (not an end-point) will never belong to the the intervals being removed at every step i, hence is in C. There are more! 3/4 and 1/13 are all in C which are not end-points of removed intervals. It is easy to observe these by considering the ternary expansion characterisation of C. Consider the ternary expansion of every x ∈ [0, 1],

∞ X ai x = = 0.a a a ... where a = 0, 1 or 2. 3i 1 2 3 3 i i=1

The decomposition of x in ternary form is not unique1. For instance, 1/3 = 0.13 = 0.022222 ...3, 2/3 = 0.23 = 0.1222 ...3 and 1 = 0.222 ...3. At the C1 stage, while removing the open interval (1/3, 2/3), we are removing all num- bers whose first digit in ternary expansion (in all possible representations) is 1. Thus, C1 has all those numbers in [0, 1] whose first digit in ternary expansion is not 1. Carrying forward this argument, we see that for each i, Ci contains all those numbers in [0, 1] with digits upto ith place, in ternary expansion, is not 1. Thus, for any x ∈ C,

∞ X ai x = = 0.a a a ... where a = 0, 2. 3i 1 2 3 3 i i=1

Lemma A.0.3. C is uncountable.

Proof. Use Cantor’s diagonal argument to show that the set of all sequences containing 0 and 2 is uncountable.

1This is true for any positional system. For instance, 1 = 0.99999 ... in decimal system APPENDIX A. CANTOR SET AND CANTOR FUNCTION 123

Cantor Function

We shall now define the Cantor function fC : C → [0, 1] as,

∞ ! ∞ X ai X ai f (x) = f = 2−i. C C 3i 2 i=1 i=1

Since ai = 0 or 2, the function replaces all 2 occurrences with 1 in the ternary expansion and we interpret the resulting number in binary system. Note, however, that the Cantor function fC is not injective. For instance, one of the representation of 1/3 is 0.0222 ...3 and 2/3 is 0.2. Under fC they are mapped to 0.0111 ...2 and 0.12, respectively, which are different representations of the same point. Since fC is same on the end-points of the removed interval, we can extend fC to [0, 1] by making it constant along the removed intervals. Alternately, one can construct the Cantor function step-by-step as we remove middle open intervals to get Ci. Consider f1 to be a function which takes the constant value 1/2 in the removed interval (1/3, 2/3) and is linear on the remaining two intervals such that f1 is continuous. In the second stage, the function f2 coincides with f1 in 1/3, 2/3, takes the constant value 1/4 and 3/4 on the two removed intervals and is linear in the remaining four intervals such that f2 is continuous. Proceeding this way we have a sequence of monotonically increasing continuous functions fk : [0, 1] → [0, 1]. −k Moreover, |fk+1(x)−fk(x)| < 2 for all x ∈ [0, 1] and fk converges uniformly to fC : [0, 1] → [0, 1].

Exercise 98. The Cantor function fC : [0, 1] → [0, 1] is uniformly continuous, 0 monotonically increasing and is differentiable a.e. and fC = 0 a.e.

Exercise 99. The function fC is not absolutely continuous.

Generalised Cantor Set

We generalise the idea behind the construction of Cantor sets to build Cantor- like subsets of [0, 1]. Choose a sequence {ak} such that ak ∈ (0, 1/2) for all k. In the first step we remove the open interval (a1, 1 − a1) from [0, 1] to get C1. Hence C1 = [0, a1] ∪ [1 − a1, 1]. Let

1 2 C1 := [0, a1] and C1 := [1 − a1, 1]. APPENDIX A. CANTOR SET AND CANTOR FUNCTION 124

1 2 i Hence, C1 = C1 ∪ C1 . Note that C1 are sets of length a1 carved out from i the end-points of C0. We repeat step one for each of the end-points of C1 of length a1a2. Therefore, we get four sets

1 2 C2 := [0, a1a2] C2 := [a1 − a1a2, a1], 3 4 C2 := [1 − a1, 1 − a1 + a1a2] C2 := [1 − a1a2, 1].

4 i i Define C2 = ∪i=1C2. Each C2 is of length a1a2. Note that a1a2 < a1. Repeating the procedure successively for each term in the sequence {ak}, k we get a sequence of sets Ck ⊂ [0, 1] whose length is 2 a1a2 . . . ak. The “generalised” Cantor set C is the intersection of all the nested Ck’s, C = ∞ 2k i ∩k=0Ck and each Ck = ∪i=1Ck. Note that by choosing the constant sequence ak = 1/3 for all k gives the Cantor set defined in the beginning of this Appendix. Similar arguments show that the generalised Cantor set C is compact. Moreover, C is non-empty, because the end-points of the closed intervals in Ck, for each k = 0, 1, 2,..., belong to C. Lemma A.0.4. For any x, y ∈ C, there is a z∈ / C such that x < z < y. Lemma A.0.5. C is uncountable.

k We show in example 2.12, that C has length 2 a1a2 . . . ak. The interesting fact about generalised Cantor set is that it can have non- zero “length”.

Proposition A.0.6. For each α ∈ [0, 1) there is a sequence {ak} ⊂ (0, 1/2) such that k lim 2 a1a2 . . . ak = α. k

Proof. Choose a1 ∈ (0, 1/2) such that 0 < 2a1−α < 1. Use similar arguments k to choose ak ∈ (0, 1/2) such that 0 < 2 a1a2 . . . ak − α < 1/k.

Generalised Cantor Function

We shall define the generalised Cantor function fC on the generalised Cantor set C. Define the function f0 : [0, 1] → [0, 1] as f0(x) = x. f0 is continuous on [0, 1]. We define f1 : [0, 1] → [0, 1] such that f is linear and continuous i on C1, and 1/2 on [a1, 1 − a1], the closure of removed open interval at first stage. We define fk : [0, 1] → [0, 1] continuous fk(0) = 0, fk(1) = 1 such that k i fk(x) = i/2 on the removed interval immediate right to Ck. APPENDIX A. CANTOR SET AND CANTOR FUNCTION 125

Theorem A.0.7. Each fk is continuous, monotonically non-decreasing and uniformly converges to some fC : [0, 1] → [0, 1].

Thus, fC being uniform limit of continuous function is continuous and is the called the generalised Cantor function. APPENDIX A. CANTOR SET AND CANTOR FUNCTION 126 Bibliography

[Pug04] Charles Chapman Pugh, Real , Undergraduate Texts in Mathematics, Springer-Verlag, Indian Reprint, 2004.

127 BIBLIOGRAPHY 128